Question

A 0.450-kg hockey puck, moving east with a speed of 5.80 m/s, has a head-on collision with a 0.900-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each puck after the collision?

Answer 1

Answer:

Explanation:

mass of first puck, m1 = 0.450 kg

initial velocity of first puck, u1 = 5.80 m/s east

mass of second puck, m2 = 0.9 kg

initial velocity of second puck, u2 = 0

Let v1 and v2 are the final velocities of the pucks after collision.

Use conservation of momentum

m1 x u1 + m2 x u2 = m1 x v1 + m2 x v2

0.45 x 5.8 + 0.9 x 0 = 0.45 x v1 + 0.9 x v2

5.8 = v1 + 2 v2 ..... (1)

As the collision is perfectly elastic, the coefficient of restitution, e = 1

Use the formula for the coefficient of restitution

$e=\frac{v_{1}-v_{2}}{u_{2}-u_{1}}$

0 - 5.8 = v1 - v2

v2 - v1 = 5.8 .... (2)

Adding equation (1) and equation (2), we get

v2 = 1.93 m/s and then v1 = - 3.87 m/s