Question

Two hockey pucks with mass 0.1 kg slide across the ice and collide. Before the collision, puck 1 is going 13 m/s to the east and puck 2 is going 18 m/s to the west. After the collision, puck 1 is going 18 m/s to the west. What is the velocity of puck 2?

Answer 1

According to the law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
0.1 x 13 + 0.1 x 18 = 0.1 x 18 + 0.1 x v2
v2 = 13 m/s
the velocity of puck 2 is 13 m/s

Answer 2

13 m/s east

Step-by-step explanation:

We can solve the problem by using the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision:

`p_i = p_f \\m u_1 + m u_2 = m v_1 + m v_2`

where

m = 0.1 kg is the mass of each puck

u1 = +13 m/s is the initial velocity of puck 1

u2 = -18 m/s is the initial velocity of puck 2 (here I assume the west direction to be the negative direction, so I put a negative sign)

v1 = -18 m/s is the final velocity of puck 1

v2 = ? is the final velocity of puck 2

Simplifying m from the formula and substituting the data, we can find the final velocity of puck 2, v2:

`v_2 = u_1 + u_2 - v_1 = +13 m/s + (-18 m/s) - (-18 m/s) = +13 m/s`

And the positive sign means that puck 2 is moving east.