A ball with a mass of # 5 kg# is rolling at #7 m/s# and elastically collides with a resting ball with a mass of #4 kg#. What are the post-collision velocities of the balls?

Question

A ball with a mass of # 5    kg# is rolling at #7    m/s# and elastically collides with a resting ball with a mass of #4  kg#. What are the post-collision velocities of the balls?

Nathan Ashcroft · Accepted Answer

I found: #0.78 and 7.78 m/s# but check my maths!

We can use conservation of momentum #vecp=mvecv# in one dimension (along #x#). So we get: #p_("before")=p_("after")# Thus: #(5*7)+(4*0)=(5*v_1)+(4*v_2)# Being an elastic collision also kinetic energy #K=1/2mv^2# is conserved: #K_("before")=K_("after")# So: #1/2*5*7^2+1/2*4*0^2=1/2*5*v_1^2+1/2*4*v_2^2# Combining the two equations yields the following result: #{(35=5v_1+4v_2),(245=5v_1^2+4v_2^2):}# From the first equation: #v_1=(35-4v_2)/5# Substitute into the second and get: #245=cancel(5)(35-4v_2)^2/cancel(25)^5+4v_2^2# #cancel(1225)=cancel(1225)-280v_2+16v_2^2+20v_2^2# #36v_2^2-280v_2=0# #v_2=0m/s# (not) and #v_2=280/36=70/9=7.78m/s# yes. Corresponding to: #v_1=7m/s# and #v_1=7/9=0.78m/s#

Amelia Adams · Answer

undefined To find the post-collision velocities, you can use the conservation of momentum and kinetic energy.

Let $ v_1 $ be the velocity of the 5 kg ball after the collision and $ v_2 $ be the velocity of the 4 kg ball after the collision.

Conservation of momentum: $ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_1 + m_2 \cdot v_2 $
$5 \cdot 7 + 4 \cdot 0 = 5 \cdot v_1 + 4 \cdot v_2 $

Conservation of kinetic energy: $ \frac{1}{2} m_1 \cdot v_{1i}^2 = \frac{1}{2} m_1 \cdot v_1^2 + \frac{1}{2} m_2 \cdot v_2^2 $
$ \frac{1}{2} \cdot 5 \cdot 7^2 = \frac{1}{2} \cdot 5 \cdot v_1^2 + \frac{1}{2} \cdot 4 \cdot v_2^2 $

Solve these two equations to find $ v_1 $ and $ v_2 $.