A ball with a mass of #2 kg# is rolling at #4 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 #2  kg# is rolling at #4   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

The velocity of the first ball is #=-1.33ms^-1#
The velocity of the second ball is #=2.67ms^-1# In an elastic collision, we have conservation of momentum and conservation of kinetic energy. The velocities before the collision are #u_1# and #u_2#. The velocities after the collision are #v_1# and #v_2#. #m_1u_1+m_2u_2=m_1v_1+m_2v_2# and #1/2m_1u_1^2+1/2m_2u_2^2=1/2m_1v_1^2+1/2m_2v_2^2# Solving the above 2 equations for #v_1# and #v_2#, we get #v_1=(m_1-m_2)/(m_1+m_2)*u_1+(2m_2)/(m_1+m_2)*u_2# and #v_2=(2m_1)/(m_1+m_2)*u_1+(m_2-m_1)/(m_1+m_2)*u_2# Taking the direction as positive #rarr^+# #m_1=2kg# #m_2=4kg# #u_1=4ms^-1# #u_2=0ms^-1# Therefore, #v_1=-2/6*4+8/6*(0)=-4/3=-1.33ms^-1# #v_2=4/6*4-2/6*(0)=8/3=2.67ms^-1# Verificaition #m_1u_1+m_2u_2=2*4+4*0=8# #m_1v_1+m_2v_2=-2*4/3+4*8/3=8#

Charles Aeschliman · Answer

undefined The post-collision velocities of the balls can be calculated using the conservation of linear momentum. The equation is $m1 * v1i + m2 * v2i = m1 * v1f + m2 * v2f$, where $m1$ and $m2$ are the masses, $v1i$ and $v2i$ are the initial velocities, and $v1f$ and $v2f$ are the final velocities.

Given:
$m1 = 2 \ kg$, $v1i = 4 \ m/s$,
$m2 = 4 \ kg$, $v2i = 0 \ m/s$ (resting ball)

Substitute values into the conservation of linear momentum equation and solve for $v1f$ and $v2f$:

$2 * 4 + 4 * 0 = 2 * v1f + 4 * v2f$

$8 = 2 * v1f + 0$

$v1f = 4 \ m/s$

Now, substitute $v1f$ into the equation:

$2 * 4 + 4 * 0 = 2 * 4 + 4 * v2f$

$8 = 8 + 0$

$v2f = 0 \ m/s$