{"id":11298,"date":"2026-07-09T14:28:03","date_gmt":"2026-07-09T14:28:03","guid":{"rendered":"https:\/\/www.myengineeringbuddy.com\/blog\/collisions\/"},"modified":"2026-07-12T04:24:30","modified_gmt":"2026-07-12T04:24:30","slug":"collisions","status":"publish","type":"post","link":"https:\/\/www.myengineeringbuddy.com\/blog\/collisions\/","title":{"rendered":"Collisions in Physics: Types, Formulas, and Solved Problems"},"content":{"rendered":"\n<div style=\"background-color:#f8f8f8; border-left:4px solid #d0d0d0; padding:12px 16px; margin-bottom:20px;\"><strong>Key Takeaways<\/strong>\n<ul>\n<li>Collisions involve a strong exchange of momentum between bodies in a short time.<\/li>\n<li>Elastic collisions conserve both momentum and kinetic energy; inelastic collisions do not.<\/li>\n<li>The coefficient of restitution (e) ranges from 0 (perfectly inelastic) to 1 (perfectly elastic).<\/li>\n<li>In oblique collisions, momentum components along the tangent remain unchanged.<\/li>\n<li>Five worked problems illustrate elastic, inelastic, and oblique collision scenarios.<\/li>\n<\/ul><\/div>\n\n<p>In this article, we will learn about collisions in detail. We will start with what collisions are and then move on to their types and then to 5 solved numerical problems. Students who want deeper support with mechanics and related topics can work with an <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/data-structures-and-algorithms-dsa\/\">online tutor for Data Structures and Algorithms (DSA)<\/a> to build the analytical problem-solving skills that underpin physics coursework.<\/p>\n\n<h2>What are Collisions?<\/h2>\n\n<p>In Physics, Collisions are the strong interactions among bodies involving an exchange of momentum in a short time interval.<\/p>\n\n<h2>Types of Collision<\/h2>\n\n<p>Based on the conservation of kinetic energy, collisions are classified into (i) Elastic Collision and (ii) Inelastic Collision.<\/p>\n\n<h3>Elastic Collision<\/h3>\n\n<p>It is the type of collision in which both the momentum and the kinetic energy are conserved.<\/p>\n\n<p>${left( {{{bf{P}}_{sy}}} right)_i} = {left( {{{bf{P}}_{sy}}} right)_f},;quad {left( {K{E_{sy}}} right)_i} = {left( {K{E_{sy}}} right)_f}$<\/p>\n\n<p>Forces involved during collisions are conservative in nature.<\/p>\n\n<p><strong>Examples:<\/strong><\/p>\n<ul>\n<li>A collision between atomic particles<\/li>\n<li>A collision between smooth billiard balls<\/li>\n<li>Collision of $alpha  &#8211; $ particle with nucleus.<\/li>\n<\/ul>\n\n<h3>Inelastic Collision<\/h3>\n\n<p>It is the collision in which the energy is not conserved. The momentum, however, is conserved.<\/p>\n\n<p>Ex: Collision between two vehicles.<\/p>\n\n<h3>Perfectly Inelastic Collisions<\/h3>\n\n<p>It is the collision in which the colliding bodies stick together and move as a single body after the collision. In a perfectly inelastic collision, the momentum remains conserved, but the loss of kinetic energy is maximum.<\/p>\n\n<p>Example: A bullet fired into the wooden block and remained embedded.<\/p>\n\n<h2>Collisions: Important Terms<\/h2>\n\n<h3>Line of Impact<\/h3>\n\n<p>The line passing through the common normal to the surface in contact during impact is called the line of impact. The force during collision act along this on both bodies.<\/p>\n\n<p style=\"text-align: center;\"><img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/fig.1-2-300x73.png\" alt=\"Diagram showing the line of impact between two colliding bodies\"><\/p>\n\n<h3>Coefficient of Restitution<\/h3>\n\n<p>Newton introduced a dimensionless parameter called the coefficient of restitution (e) to measure the elasticity of collision. It is defined as the ratio of the relative velocity of separation to the relative velocity of approach of the two colliding bodies.<\/p>\n\n<p>$e = frac{{{rm{Relativetext{-}velocitytext{-}oftext{-}separation}}}}{{{rm{Relativetext{-}velocitytext{-}oftext{-}approach}}}}$<\/p>\n\n<p>$ Rightarrow quad left( {{{bf{v}}_2} &#8211; {{bf{v}}_1}} right) = eleft( {{{bf{u}}_1} &#8211; {{bf{u}}_2}} right)$<\/p>\n\n<p>This formula is applied along the line of impact. Here, the velocities mentioned in the expression should be taken along the line of impact.<\/p>\n\n<p>For perfectly elastic collision, e = 1. For an inelastic collision, 0 &lt; e &lt; 1. For perfectly inelastic collision e = 0.<\/p>\n\n<h2>Head-On Elastic Collision<\/h2>\n\n<p>Let two bodies of masses ${m_1}$ and ${m_2}$ moving with initial velocities ${{bf{u}}_1}$ and ${{bf{u}}_2}$ $({{bf{u}}_1} &gt; {{bf{u}}_2})$. After collision two bodies will move with velocity ${{bf{v}}_1}$ and ${{bf{v}}_2}$.<\/p>\n\n<p>From the law of conservation of linear momentum,<\/p>\n\n<p>${m_1}{{bf{u}}_1} + {m_2}{{bf{u}}_2} = {m_1}{{bf{v}}_1} + {m_2}{{bf{v}}_2}$ &nbsp;&nbsp;&nbsp;\u2026(1)<\/p>\n\n<p>And by the definition of coefficient of restitution,<\/p>\n\n<p>$({{bf{v}}_2} &#8211; {{bf{v}}_1}) = e({{bf{u}}_1} &#8211; {{bf{u}}_2})$ &nbsp;&nbsp;&nbsp;\u2026(2)<\/p>\n\n<p>Solving (1) &amp; (2), for ${{bf{v}}_1}$ and ${{bf{v}}_2}$,<\/p>\n\n<p>${{{bf{v}}_1} = left( {frac{{{m_1} &#8211; e{m_2}}}{{{m_1} + {m_2}}}} right){{bf{u}}_1} + left( {frac{{(1 + e){m_2}}}{{{m_1} + {m_2}}}} right){{bf{u}}_2}}$<\/p>\n\n<p>${{{bf{v}}_2} = left( {frac{{(1 + e){m_1}}}{{{m_1} + {m_2}}}} right){{bf{u}}_1} + left( {frac{{{m_2} &#8211; e{m_1}}}{{{m_1} + {m_2}}}} right){{bf{u}}_2}}$<\/p>\n\n<p>Loss in kinetic energy of system:<\/p>\n\n<p>$Delta KE = {left( {K{E_{sy}}} right)_i} &#8211; {left( {K{E_{sy}}} right)_f}$<\/p>\n\n<p>$Delta KE = frac{1}{2}left( {frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}} right){left| {{{bf{u}}_1} &#8211; {{bf{u}}_2}} right|^2}(1 &#8211; {e^2})$<\/p>\n\n<p>In the case of perfectly inelastic collision, e = 0. Loss in KE of system is $Delta KE = frac{1}{2}left( {frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}} right){left( {left| {{{bf{u}}_1} &#8211; {{bf{u}}_2}} right|} right)^2}$<\/p>\n\n<p>If two bodies are approaching each other, then the loss in KE of the system is maximum. Students exploring how predictive modelling connects to energy-loss calculations may find it useful to consult a <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/machine-learning\/\">machine learning tutor<\/a> for applied computational methods.<\/p>\n\n<p>${left[ {{{(Delta KE)}_{sys}}} right]_{max }} = frac{1}{2}left( {frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}} right){({u_1} + {u_2})^2}$<\/p>\n\n<h2>Oblique Collisions (or Collisions in Two Dimensions)<\/h2>\n\n<p>(1) A pair of equal and opposite impulses act along a common normal direction. Hence linear momentum of individual particles changes along the common normal direction.<\/p>\n\n<p>(2) No component of impulse acts along the common tangent direction. Hence linear momentum (or) linear velocity of individual particles remains unchanged along this direction.<\/p>\n\n<p>(3) Net impulse on the particles is zero during the collision. Hence, the net momentum of both the particles remains conserved before and after collision in any direction.<\/p>\n\n<p>(4) The equation for the coefficient of restitution can be applied along the common normal direction or the line of impact. Engineers who model two-dimensional impact scenarios often use <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/autocad\/\">AutoCAD tutoring<\/a> to visualise force vectors and geometric relationships accurately.<\/p>\n\n<p style=\"text-align: center;\"><img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/fig.2-1-300x52.png\" alt=\"Diagram illustrating oblique collision between two particles in two dimensions\"><\/p>\n\n<p>From law of conservation of linear momentum along x-axis,<\/p>\n\n<p>${m_1}{u_1}cos {theta _1} + {m_2}{u_2}cos {theta _2} = {m_1}{v_1}cos {beta _1} + {m_2}{v_2}cos {beta _2}$<\/p>\n\n<p>Along y-axis,<\/p>\n\n<p>${m_1}{u_1}sin {theta _1} + {m_2}{u_2}sin {theta _2} = {m_1}{v_1}sin {beta _1} + {m_2}{v_2}sin {beta _2}$<\/p>\n\n<p>Using the velocity of separation = e (velocity of approach) along the line of impact,<\/p>\n\n<p>$left( {{v_2}cos {beta _2} &#8211; {v_1}cos {beta _1}} right) = eleft( {{u_1}cos {theta _1} &#8211; {u_2}cos {theta _2}} right)$<\/p>\n\n<p>For students who want to explore how agile learning frameworks can support mastering multi-step physics derivations, this overview of <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/agile-tutoring-flexible-student-centered-learning-support\/\">agile tutoring and flexible student-centred learning support<\/a> is worth reading.<\/p>\n\n<h2>Solved Problems on Collisions<\/h2>\n\n<h3>Problem 1: Elastic Head-On Collision<\/h3>\n\n<p><strong>Two bodies of masses ${m_1}$ and ${m_2}$ are moving with velocities of 1 m\/s and 3 m\/s, respectively, in opposite directions. If the bodies undergo one-dimensional elastic collision, the body of mass ${m_1}$ comes to rest. Find the ratio of ${m_1}$ and ${m_2}$.<\/strong><\/p>\n\n<p><strong>Answer<\/strong><\/p>\n\n<p>Let bodies ${m_1}$ and ${m_2}$ as a system<\/p>\n\n<p style=\"text-align: center;\"><img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/fig.3-1-300x77.png\" alt=\"Diagram of elastic head-on collision between two bodies m1 and m2\"><\/p>\n\n<p>Since, $({{bf{F}}_{ext}}) = 0,;,,{({{bf{P}}_{sys}})_i} = {({{bf{P}}_{sys}})_f}$<\/p>\n\n<p>$ Rightarrow quad {m_1}{{bf{v}}_1} + {m_2}{{bf{v}}_2} = {m_1}{bf{v}}_1^{bf{&#8216;}} + {m_2}{bf{v}}_2^{bf{&#8216;}}$<\/p>\n\n<p>$ Rightarrow quad {m_1}(1,{rm{m\/s}})hat i + {m_2}(3,{rm{m\/s}})( &#8211; hat i) = {m_1} times 0 + {m_2}{bf{v}}_2^{bf{&#8216;}}$<\/p>\n\n<p>$ Rightarrow quad {m_1}hat i &#8211; 3{m_2}hat i = {m_1}{bf{v}}_2^{bf{&#8216;}}$ &nbsp;&nbsp;&nbsp;\u2026(1)<\/p>\n\n<p>Using, ${bf{v}}_2^{bf{&#8216;}} &#8211; {bf{v}}_1^{bf{&#8216;}} = e,({{bf{v}}_1} &#8211; {{bf{v}}_2})$<\/p>\n\n<p>$ Rightarrow quad {bf{v}}_2^{bf{&#8216;}} &#8211; 0 = 1,left( {1hat i &#8211; 3( &#8211; hat i)} right)$<\/p>\n\n<p>$ Rightarrow quad {bf{v}}_2^{bf{&#8216;}} = 4,hat i$ &nbsp;&nbsp;&nbsp;\u2026(2)<\/p>\n\n<p>From (1) &amp; (2),<\/p>\n\n<p>${m_1}hat i &#8211; 3{m_2}hat i = 4{m_2}hat iquad  Rightarrow quad {m_1} = 7{m_2}$<\/p>\n\n<p>$ Rightarrow quad frac{{{m_1}}}{{{m_2}}} = frac{7}{1}$<\/p>\n\n<h3>Problem 2: Inelastic Head-On Collision<\/h3>\n\n<p><strong>Ball 1 collides with another identical ball 2 at rest as shown in the figure. For what value of the coefficient of restitution e, the velocity of the second ball becomes two times that of the first ball after the collision.<\/strong><\/p>\n\n<p style=\"text-align: center;\"><img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/fig.4-300x60.png\" alt=\"Diagram of ball 1 colliding with stationary ball 2 in a head-on inelastic collision\"><\/p>\n\n<p><strong>Answer<\/strong><\/p>\n\n<p>Since both balls are identical therefore, ${m_1} = {m_2}$.<\/p>\n\n<p>Let initial speeds of balls 1 and 2 are ${v_1}$ and ${v_2}$ and after collision their speeds are $v_1^{bf{&#8216;}}$ and $v_2^{bf{&#8216;}}$.<\/p>\n\n<p style=\"text-align: center;\"><img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/fig.5-300x53.png\" alt=\"Diagram showing velocities before and after inelastic head-on collision of two identical balls\"><\/p>\n\n<p>Using momentum conservation ; $sumlimits_{}^{} {{{bf{F}}_{ext}}}  = 0$.<\/p>\n\n<p>${m_1}{{bf{v}}_1} + {m_2}{{bf{v}}_2} = {m_1}{bf{v}}_1^{bf{&#8216;}} + {m_2}{bf{v}}_2^{bf{&#8216;}}$<\/p>\n\n<p>Since, ${m_1} = {m_2},;,,{v_2} = 0,;,,{bf{v}}_2^{bf{&#8216;}} = 2{bf{v}}_1^{bf{&#8216;}}$<\/p>\n\n<p>${{{bf{v}}_1} + 0 = {bf{v}}_1^{bf{&#8216;}} + {m_2}{bf{v}}_2^{bf{&#8216;}}}$ &nbsp;&nbsp;&nbsp;\u2026(1)<\/p>\n\n<p>Using, ${bf{v}}_2^{bf{&#8216;}} &#8211; {bf{v}}_1^{bf{&#8216;}} = eleft( {{{bf{v}}_1} &#8211; {{bf{v}}_2}} right)$<\/p>\n\n<p>$ Rightarrow quad 2{bf{v}}_1^{bf{&#8216;}} &#8211; {bf{v}}_1^{bf{&#8216;}} = eleft( {{{bf{v}}_1} &#8211; 0} right)$<\/p>\n\n<p>$ Rightarrow quad {bf{v}}_1^{bf{&#8216;}} = e,{{bf{v}}_1}$<\/p>\n\n<p>Now, from (1),<\/p>\n\n<p>${bf{v}}_1^{bf{&#8216;}} = e,3{bf{v}}_1^{bf{&#8216;}}quad  Rightarrow quad e = frac{1}{3}$<\/p>\n\n<h3>Problem 3: Inelastic Oblique Collision<\/h3>\n\n<p><strong>After a perfectly inelastic collision between two identical particles moving at the same speed but in different directions, the speed of the combined particle becomes half the initial speed of either particle. The angle between the velocities of the two before the collision is \u2026<\/strong><\/p>\n\n<p><strong>Answer<\/strong><\/p>\n\n<p>System: both particles; Here $left( {{{bf{F}}_{ext}}} right)$ acting on the system = 0<\/p>\n\n<p>${{{bf{P}}_1} + {{bf{P}}_2} = {bf{P}}}$<\/p>\n\n<p>$m{{bf{v}}_1} + m{{bf{v}}_2} = 2m{{bf{v}}_f}$ &nbsp;&nbsp;&nbsp;\u2026(1)<\/p>\n\n<p>Here, $left| {{{bf{v}}_1}} right| = left| {{{bf{v}}_2}} right|$; let ${v_1} = {v_2} = {v_0}$ then ${v_f} = frac{{{v_0}}}{2}$<\/p>\n\n<p>From (1),<\/p>\n\n<p>${(m{v_0})^2} + {(m{v_0})^2} + 2m{v_0}m{v_0}cos theta  = {left( {2mfrac{{{v_0}}}{2}} right)^2}$<\/p>\n\n<p>$ Rightarrow quad {m^2}v_0^2 + {m^2}v_0^2 + 2{m^2}v_0^2cos theta  = {(2m)^2}frac{{v_0^2}}{4}$<\/p>\n\n<p>$ Rightarrow quad v_0^2 + v_0^2 + 2v_0^2cos theta  = v_0^2$<\/p>\n\n<p>$ Rightarrow quad 2v_0^2cos theta  =  &#8211; v_0^2$<\/p>\n\n<p>$ Rightarrow quad cos theta  =  &#8211; frac{1}{2}quad  Rightarrow quad theta  = {120^o}$<\/p>\n\n<h3>Problem 4: Energy Loss in a Head-On Collision<\/h3>\n\n<p><strong>A block of mass 0.5 kg is moving with a speed of 2.0 m\/s on a smooth surface. It strikes a stationary mass of 1.00 kg head-on, and then they move together as a single body. Find the energy loss during the collision.<\/strong><\/p>\n\n<p><strong>Answer<\/strong><\/p>\n\n<p>System: both mass;<\/p>\n\n<p>$sumlimits_{}^{} {{{bf{F}}_{ext}}}  = 0,,;,,{left( {{{bf{P}}_{sys}}} right)_i} = {left( {{{bf{P}}_{sys}}} right)_f}$<\/p>\n\n<p>$ Rightarrow quad {m_1}{{bf{v}}_1} + {m_1}{{bf{v}}_2} = ({m_1} + {m_2}){{bf{v}}_f}$<\/p>\n\n<p>$ Rightarrow quad (0.5 times 2) + (1 times 0) = (1.5){v_f}$<\/p>\n\n<p>$ Rightarrow quad {v_f} = frac{2}{3}{rm{m\/s}}$<\/p>\n\n<p>Now, loss in KE of the system:<\/p>\n\n<p>$Delta KE = left( {frac{1}{2}{m_1}v_1^2 + frac{1}{2}{m_2}v_2^2} right) &#8211; left( {frac{1}{2}{m_1}v_f^2 + frac{1}{2}{m_2}v_f^2} right)$<\/p>\n\n<p>$ = frac{1}{2} times (0.5){(2)^2} &#8211; frac{1}{2} times 1.5 times {left( {frac{2}{3}} right)^2} = 0.67,J$<\/p>\n\n<p>For a broader look at how online and classroom-based instruction compare when tackling quantitative subjects, see this comparison of <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/online-vs-classroom-machine-learning-courses-which-delivers-better-results\/\">online vs classroom machine learning courses and which delivers better results<\/a>.<\/p>\n\n<h3>Problem 5: Inelastic Oblique Collisions of Three Billiard Balls<\/h3>\n\n<p><strong>Two billiard balls of the same size (radius r) and the same mass are in contact on a billiard table. A third ball of the same size and mass strikes them symmetrically and remains at rest after the impact. Find the coefficient of restitution.<\/strong><\/p>\n\n<p style=\"text-align: center;\"><img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/fig.6-300x90.png\" alt=\"Diagram of a third billiard ball striking two billiard balls in contact symmetrically\"><\/p>\n\n<p><strong>Answer<\/strong><\/p>\n\n<p>System: all three balls as a system.<\/p>\n\n<p>$sumlimits_{}^{} {{{bf{F}}_{ext}}}  = 0,,;,,{left( {{{bf{P}}_{sys}}} right)_i} = {left( {{{bf{P}}_{sys}}} right)_f}$<\/p>\n\n<p>${left[ {{{left( {{{bf{P}}_{sys}}} right)}_x}} right]_i} = {left[ {{{left( {{{bf{P}}_{sys}}} right)}_x}} right]_f}$<\/p>\n\n<p>$ Rightarrow quad mu = mvcos theta  + mvcos theta $<\/p>\n\n<p>$ Rightarrow quad u = 2vcos theta $<\/p>\n\n<p>$ Rightarrow quad u = 2v times frac{{sqrt 3 }}{2}quad  Rightarrow quad u = sqrt 3 v$<\/p>\n\n<p>Now using the equation, velocity of separation = e (velocity of approach)<\/p>\n\n<p>Along the line of impact,<\/p>\n\n<p style=\"text-align: center;\"><img class=\"lazyload\" decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAAAAACH5BAEKAAEALAAAAAABAAEAAAICTAEAOw==\" data-orig-src=\"https:\/\/myengineeringbuddy.com\/wp-content\/uploads\/2022\/07\/fig.7-300x177.png\" alt=\"Diagram showing line of impact and velocity components for three billiard ball collision\"><\/p>\n\n<p>$v &#8211; 0 = e(ucos {30^o} &#8211; 0)$<\/p>\n\n<p>$ Rightarrow quad v = e cdot sqrt 3 v cdot frac{{sqrt 3 }}{2}quad  Rightarrow quad e = frac{2}{3}$<\/p>\n\n<p>Learners who want to understand how spatial and geometric reasoning applies to engineering problems may benefit from reading about <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/geomatics-tutoring-spatial-learning-homework-help\/\">geomatics tutoring and spatial learning<\/a>. Those looking to sharpen their abstract mathematical foundations can also explore working with a <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/abstract-algebra\/\">tutor for abstract algebra<\/a>. For students in actuarial or risk-related programmes where probability and collision modelling intersect, connecting with an <a href=\"https:\/\/www.myengineeringbuddy.com\/subject\/actuarial-science\/\">actuarial science expert<\/a> can provide valuable applied context. Platforms that offer coding practice alongside physics problem-solving are reviewed in this article on <a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/hackerrank-reviews-alternatives-pricing-offerings\/\">HackerRank reviews, alternatives, pricing, and offerings<\/a>.<\/p>\n\n<p>Related: <a href=\"https:\/\/www.myengineeringbuddy.com\/work-done-in-physics\/\">Work done in physics<\/a> | <a href=\"https:\/\/www.myengineeringbuddy.com\/torque\/\">Torque<\/a> | <a href=\"https:\/\/www.myengineeringbuddy.com\/homework-help\/physics-homework-help\/\">Physics Homework Help<\/a><\/p>\n\n<h2>Related Reading<\/h2>\n<ul>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/collegeessay-reviews-alternatives-pricing-offerings-in-2025\/\">CollegeEssay Reviews, Alternatives, Pricing and Offerings in 2025<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/mastering-geography-with-tutoring-a-guide-for-students\/\">Mastering Geography with Tutoring: A Guide for Students<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/education-tutoring-personalized-learning-and-online-teaching-for-students\/\">Education Tutoring: Personalized Learning and Online Teaching for Students<\/a><\/li>\n<li><a href=\"https:\/\/www.myengineeringbuddy.com\/blog\/book-creator-reviews-alternatives\/\">Book Creator Reviews and Alternatives<\/a><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Key Takeaways Collisions involve a strong exchange of momentum between  [&#8230;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-11298","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11298","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/comments?post=11298"}],"version-history":[{"count":1,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11298\/revisions"}],"predecessor-version":[{"id":12174,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/posts\/11298\/revisions\/12174"}],"wp:attachment":[{"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/media?parent=11298"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/categories?post=11298"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.myengineeringbuddy.com\/blog\/wp-json\/wp\/v2\/tags?post=11298"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}