A 0.471-kg hockey puck, moving east with a speed of 8.17 m/s, has a head-on collision with a 0.795-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed of the 0.795-kg puck after the collision?
Given:
M1 = 0.795kg, V1 = 0.
M2 = 0.471kg, V2 = 8.17 m/s.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.795*0 + 0.471*8.17 = 0.795*V3 + 0.471*V4,
Eq1: 0.795*V3 + 0.471*V4 = 3.85.
Conservation of KE Eq:
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (0(M1-M2) + 2*0.471*8.17)/(0.795+0.471) = 6.08 m/s = Velocity of M1 after the collision.
To determine the speed of the 0.795-kg puck after the collision, we can apply the law of conservation of momentum.
The law of conservation of momentum states that the total momentum of an isolated system remains constant before and after a collision. In other words, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision, the momentum of the system is given by:
Momentum before = (mass of first puck) * (velocity of first puck) + (mass of second puck) * (velocity of second puck)
Momentum before = (0.471 kg) * (8.17 m/s) + (0.795 kg) * (0 m/s)
Since the second puck is initially at rest, its velocity is 0 m/s.
Momentum before = (0.471 kg) * (8.17 m/s) + (0.795 kg) * (0 m/s)
= 3.871 kg·m/s
Let's assume the speed of the 0.795-kg puck after the collision is v2.
After the collision, the momentum of the system is given by:
Momentum after = (mass of first puck) * (velocity of first puck after collision) + (mass of second puck) * (velocity of second puck after collision)
Since the collision is perfectly elastic, the relative velocity of the two pucks will remain the same but change direction. Therefore, the velocity of the first puck after the collision will be (v2 - 8.17) m/s, and the velocity of the second puck after the collision will be (v2 - 0) m/s.
Momentum after = (0.471 kg) * (v2 - 8.17) m/s + (0.795 kg) * (v2 - 0) m/s
According to the law of conservation of momentum:
Momentum before = Momentum after
(0.471 kg) * (8.17 m/s) + (0.795 kg) * (0 m/s) = (0.471 kg) * (v2 - 8.17) m/s + (0.795 kg) * (v2 - 0) m/s
3.871 kg·m/s = (0.471 kg) * (v2 - 8.17) m/s + (0.795 kg) * (v2 - 0) m/s
Now, we can solve this equation to find the value of v2.
3.871 kg·m/s = (0.471 kg) * v2 - (0.471 kg) * 8.17 m/s + (0.795 kg) * v2
3.871 kg·m/s = (0.471 kg + 0.795 kg) * v2 - (0.471 kg) * 8.17 m/s
3.871 kg·m/s = (1.266 kg) * v2 - (0.471 kg) * 8.17 m/s
3.871 kg·m/s + (0.471 kg) * 8.17 m/s = (1.266 kg) * v2
Now, we can isolate v2:
v2 = (3.871 kg·m/s + (0.471 kg) * 8.17 m/s) / (1.266 kg)
Calculating this, we get:
v2 ≈ 8.792 m/s
Therefore, the speed of the 0.795-kg puck after the collision is approximately 8.792 m/s.