Question

Puck 1 (0.5 kg) travels with velocity 40 m/s to the right when it collides with puck 2 (1 kg) which is initially at rest. After the collision, puck 1 moves with a velocity of 0 m/s. Assume that no external forces are present and therefore the momentum for the system of pucks is conserved. What is the final velocity (in m/s) of puck 2 after the collision

Answer 1

The final velocity of puck 2 after the collision is 20 m/s .

Step-by-step explanation:

Let us assume , right direction is positive and left direction is negative .

Given :

Mass of puck 1 and 2 of 0.5 kg and 1 kg respectively .

Initial velocity of puck 1 is 40 m/s .

Initial velocity of puck 2 is 0 m/s .

Final velocity of puck 1 is 0 m/s .

Let , us assume final velocity of puck 2 is
`v_2` .

Therefore , by conservation of momentum :

`0.5* 40+1* 0=0.5* 0+1* v_2\\v_2=20\ m/s`

Therefore , the final velocity (in m/s) of puck 2 after the collision is 20 m/s .

Answer 2

20 m/s

Step-by-step explanation:

mass of puck 1, m1 = 0.5 kg

initial velocity, u1 = 40 m/s

mass of puck 2, m2 = 1 kg

initial velocity of puck 2, u2 = 0 m/s

final velocity of puck 1, v1 = 0 m/s

Let the final velocity of puck 2 is v2.

As there is not external force acting on the system, the momentum of the system is conserved.

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

0.5 x 40 + 1 x 0 = 0.5 x 0 + 1 x v2

20 = v2

v2 = 20 m/s

Thus, the velocity of puck 2 after the collision is 20 m/s.