A ball with a mass of # 5 kg# is rolling at #6 m/s# and elastically collides with a resting ball with a mass of #3 kg#. What are the post-collision velocities of the balls?

Question

A ball with a mass of # 5    kg# is rolling at #6   m/s# and elastically collides with a resting ball with a mass of #3  kg#. What are the post-collision velocities of the balls?

Layla Ahmed · Accepted Answer

#color(red)(v_1^'=3/2 m/s)#
#color(red)(v_2^'=15/2)# #"momentums before collision"# #P_1=m_2*v_1=5*6=30" " kg*m/s# #P_2=m_2*v_2=3*0=0# #Sigma vec P_b=vecP_1+vec P_2" total momentum before collision"# #Sigma vec P_b=30+0=30 " "kg*m/s# #"total momentum after collision must be "30 " "kg.m/s# #"momentums after collision"# #vec P_1^'=m_1*v_1^'=5*v_1^'# #vec P_2^'=m_2*v_2^'=3*v_2^'# #Sigma P_a=vec P_1^'+vec P_2^'# #Sigma P_a=5*v_1^'+3*v_2^'# #Sigma vec P_b=Sigma vec P_a# #30=5*v_1^'+3*v_2^'" "(1)"# #v_1+v_1^'=v_2+v_2^'# #6+v_1^'=0+v_2^'# #v_2^'=6+v_1^'" (2)"# #"use the equation (1)"# #30=5v_1^'+3(6+v_1^')# #30=5v_1^'+18+3v_1^'# #30-18=8v_1^'# #12=8v_1^2# #v_1^'=12/8# #color(red)(v_1^'=3/2 m/s)# #"use (2)"# #v_2^'=6+v_1^'# #v_2^'=6+1,5# #color(red)(v_2^'=15/2)#

Axel Acevedo · Answer

undefined To find the post-collision velocities of the balls, we can use the principle of conservation of momentum and the conservation of kinetic energy:

1. Calculate the total momentum before the collision.
2. Apply the conservation of momentum to find the velocities of the balls after the collision.
3. Apply the conservation of kinetic energy to check if the collision is elastic.

Let $ v_1 $ be the velocity of the 5 kg ball after the collision, and $ v_2 $ be the velocity of the 3 kg ball after the collision.

Conservation of momentum: $ m_1v_{1i} + m_2v_{2i} = m_1v_1 + m_2v_2 $

Conservation of kinetic energy: $ \frac{1}{2}m_1v_{1i}^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 $

Given:
$ m_1 = 5 $ kg, $ v_{1i} = 6 $ m/s
$ m_2 = 3 $ kg, $ v_{2i} = 0 $ m/s (resting)

Conservation of momentum: $ 5(6) + 3(0) = 5v_1 + 3v_2 $
$ 30 = 5v_1 + 3v_2 $ ...........(1)

Conservation of kinetic energy: $ \frac{1}{2}(5)(6)^2 = \frac{1}{2}(5)v_1^2 + \frac{1}{2}(3)v_2^2 $
$ 90 = 15v_1^2 + \frac{3}{2}v_2^2 $ ...........(2)

Solve equations (1) and (2) simultaneously to find $ v_1 $ and $ v_2 $.