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8.6 Collisions of Point Masses in Two Dimensions

Splitting momentum conservation into x and y components to recover speed and scattering angle after a glancing hit between point masses.

College Physics 2e, 8.6 Collisions of Point Masses in Two Dimensions

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5 exercises

8.6 Exercise 1

Two cars of equal mass approach an intersection. Car A is moving east at a speed of 45 m/s. Car B is moving south at a speed of 35 m/s. They collide inelastically and stick together after the collision, moving as one object. Which of the following statements is true about the center-of-mass velocity of this system?

  1. The center-of-mass velocity will decrease after the collision as a result of lost…
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8.6 Exercise 4

Two identical pucks collide on an air hockey table. One puck was originally at rest. (a) If the incoming puck has a speed of 6.00 m/s and scatters to an angle of 30.0º\text{30} . 0º,what is the velocity (magnitude and direction) of the second puck? (You may use the result that θ1−θ2=90º\theta_{1} - \theta_{2} = \text{90}º for elastic collisions of objects that have identical masses.) (b) Confirm that the collision is elastic.

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8.6 Exercise 6

A 3000-kg cannon is mounted so that it can recoil only in the horizontal direction. (a) Calculate its recoil velocity when it fires a 15.0-kg shell at 480 m/s at an angle of 20.0º\text{20} . 0º above the horizontal. (b) What is the kinetic energy of the cannon? This energy is dissipated as heat transfer in shock absorbers that stop its recoil. (c) What happens to the vertical component of momentum that is imparted…

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8.6 Exercise 8

Professional Application

Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in Chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei (4He)\left(^{4} \text{He}\right) from gold-197 nuclei (197Au)\left(^{\text{197}} \text{Au}\right). The energy of the incoming helium nucleus was 8.00×10−13J8.00 \times \text{10}^{- \text{13}} \text{J}, and the masses of the helium and gold nuclei were 6.68×10−27 kg6.68 \times \text{10}^{- \text{27}} \text{ kg} and 3.29×10−25kg3.29 \times \text{10}^{- \text{25}} \text{kg}, respectively (note that their mass ratio is 4 to 197). (a) If a helium n…

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8.6 Exercise 10

Starting with equations m1v1=m1 v1′cosθ1+m2 v2′cosθ2m_{1} v_{1} = m_{1} v_{1}^{'} \text{cos} \theta_{1} + m_{2} v_{2}^{'} \text{cos} \theta_{2} and 0=m1 v1′sinθ1+m2 v2′sinθ20 = m_{1} v_{1}^{'} \text{sin} \theta_{1} + m_{2} v_{2}^{'} \text{sin} \theta_{2} for conservation of momentum in the xx- and yy-directions and assuming that one object is originally stationary, prove that for an elastic collision of two objects of equal masses,

1 2 mv 1 2 = 1 2 mv ′ 1 2 + 1 2 mv ′ 2 2 + mv′ 1 v ′ 2 cos ( θ 1 − θ 2 ) \begin{matrix}\frac{1}{2} \left(\text{mv}_{1}\right)^{2} = \\ \frac{1}{2} \left(\left(\text{mv} '\right)_{1}\right)^{2} + \frac{1}{2} \left(\left(\text{mv} '\right)_{2}\right)^{2} \\ + \left(\text{mv} '\right)_{1} \left(v '\right)_{2} \text{cos} \left(\theta_{1} - \theta_{2}\right)\end{matrix}

as discussed in the text.

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Adapted from College Physics © 2022 Rice University. Licensed under CC BY 4.0. Sources: College Physics 2e, College Physics for AP® Courses 2e. Imported at revision f98d7a7.