A 45kg student is riding on a 7.0kg skateboard with a velocity of +4.0m/s. The student jumps off the board with a velocity of -1.0m/s. Find the velocity of the skateboard after the student jumped off.
A 5.0kg ball moving at 6.0m/s strikes a 3.0kg ball initially at rest. The 5.0kg ball continues moving in the same direction at 2.0m/s after the collision. Find the velocity of the 3.0kg ball.This is a conservation of momentum problem. The total momentum of the system (student + skateboard) before the jump is equal to the total momentum after the jump.
Before the jump:
Total mass = 45 kg + 7 kg = 52 kg
Total momentum = (52 kg) x (4.0 m/s) = 208 kg*m/s
After the jump:
The mass of the skateboard remains the same, but the student is no longer on it. So the total mass is now just 7 kg. Let's call the velocity of the skateboard after the jump v:
Total mass = 7 kg
Total momentum = (7 kg) x v
According to conservation of momentum:
Total momentum before = Total momentum after
(52 kg) x (4.0 m/s) = (7 kg) x v
Solving for v, we get:
v = (52 kg x 4.0 m/s) / 7 kg
v = 29.71 m/s
So the velocity of the skateboard after the student jumped off is approximately 29.71 m/s. Note that this assumes no external forces (such as friction) act on the system.
This is also a conservation of momentum problem. The total momentum before the collision is equal to the total momentum after the collision.A 5.0kg ball moving at 6.0m/s strikes a 3.0kg ball initially at rest. The 5.0kg ball continues moving in the same direction at 2.0m/s after the collision. Find the velocity of the 3.0kg ball.
Thank you.This is also a conservation of momentum problem. The total momentum before the collision is equal to the total momentum after the collision.
Before the collision:
Total momentum = (5.0 kg) x (6.0 m/s) = 30 kg m/s (since the smaller ball is at rest)
After the collision:
Let's call the velocity of the 5.0 kg ball after the collision v1 and the velocity of the 3.0 kg ball after the collision v2.
Total momentum = (5.0 kg) x (2.0 m/s) + (3.0 kg) x v2
According to conservation of momentum:
Total momentum before = Total momentum after
30 kg m/s = (5.0 kg) x (2.0 m/s) + (3.0 kg) x v2
Solving for v2, we get:
v2 = (30 kg m/s - (5.0 kg) x (2.0 m/s)) / (3.0 kg)
v2 = 6.7 m/s (rounded to one decimal place)
So the velocity of the 3.0 kg ball after the collision is approximately 6.7 m/s, in the same direction as the 5.0 kg ball.
You're welcome! If you have any other physics questions, feel free to ask.Thank you.