A ball with a mass of #2 kg# is rolling at #12 m/s# and elastically collides with a resting ball with a mass of #1 kg#. What are the post-collision velocities of the balls?

Question

A ball with a mass of #2  kg# is rolling at #12    m/s# and elastically collides with a resting ball with a mass of #1 kg#. What are the post-collision velocities of the balls?

William Audy · Accepted Answer

#v_1=4 m/sec#
#v_2=16 m/sec# #2*12+1*0=2*v_1+1*v_2# #24=2*v_1+v_2# #12+v_1=v_2# #24=2*v_1+12+v_1# #24-12=3*v1# #12=3*v_1# #v_1=4 m/sec# #12+4=v_2# #v_2=16 m/sec#

Leo Abeyta · Answer

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

Let $v_1$ and $v_2$ be the velocities of the 2 kg ball and 1 kg ball respectively after the collision.

Using the conservation of momentum:

$m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2$

where:
$m_1 = 2$ kg (mass of the first ball)
$m_2 = 1$ kg (mass of the second ball)
$u_1 = 12$ m/s (initial velocity of the first ball)
$u_2 = 0$ m/s (initial velocity of the second ball)

Plugging in the values:

$2 \times 12 + 1 \times v_2 = 2 \times v_1 + 1 \times 0$

$24 + v_2 = 2v_1$

Now, using the conservation of kinetic energy:

$0.5m_1u_1^2 + 0.5m_2u_2^2 = 0.5m_1v_1^2 + 0.5m_2v_2^2$

Plugging in the values:

$0.5 \times 2 \times 12^2 + 0.5 \times 1 \times 0^2 = 0.5 \times 2 \times v_1^2 + 0.5 \times 1 \times v_2^2$

$144 = 2v_1^2 + 0.5v_2^2$

We now have a system of equations:

$24 + v_2 = 2v_1$ ...(1)

$144 = 2v_1^2 + 0.5v_2^2$ ...(2)

Solving these equations simultaneously will give us the values of $v_1$ and $v_2$.