Question

On a frictionless horizontal air table, puck A (with mass 0.252 kg ) is moving toward puck B (with mass 0.374 kg ), which is initially at rest. After the collision, puck A has velocity 0.115 m/s to the left, and puck B has velocity 0.650 m/s to the right. Calculate ΔK, the change in the total kinetic energy of the system that occurs during the collision.

Answer 1

As we know that kinetic energy is given as

$KE = \frac{1}{2}mv^2$

Here we can find the initial speed of puck A by momentum conservation

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$0.252\times v + 0 = 0.252\times (-0.115) + 0.374\times ( 0.650)$

$v = 0.850 m/s$

now here we will have initial kinetic energy of the mass is given as

$KE_i = \frac{1}{2}m_1v_1^2$

$KE_i = \frac{1}{2}(0.252)(0.850^2) = 0.091 J$

$KE_f = \frac{1}{2}(0.252)(-0.115)^2 + \frac{1}{2}(0.374)(0.650^2)$

$KE_f = 0.081 J$

now loss of energy is given as

$KE_i - KE_f = 0.091 - 0.081 = 0.010 J$

Answer 2

The change in the total kinetic energy of the system that occurs during the collision is 0.010Joules

According to the law of collision;

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Substituting the given parameters into the equation to get the velocity of puck A, we have;

$0.252u_1+0=0.252(-0.115)+(0.650)(0.374)\\0.252u_1=-0.02898 + 0.2431\\0.252u_1=0.21412\\u_1=\frac{0.21412}{0.252}\\u_1= 0.85m/s$

Change in kinetic energy = kinetic energy before the collision - kinetic energy after the collision

Change in kinetic energy = $\frac{1}{2}(0.252)(0.115)^2+\frac{1}{2}(0.374)(0.650)^2 - \frac{1}{2}(0.252)(0.86)$

Change in kinetic energy = $0.091 - 0.081$

Change in kinetic energy = 0.010 Joules

Hence the change in the total kinetic energy of the system that occurs during the collision is 0.010Joules

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