Question

A ball of mass 0.32 kg and a velocity of 5.1 m/s collides head-on with a ball of mass 0.75 kg that is initially at rest. no external forces act on the balls. after the collision, the velocity of the ball which was originally at rest is 2.38 m/s. what is the velocity of the 0.32 kg ball after the collision? remember to include your data, equation, and work when solving this problem.

Answer 1

To solve this problem, we will use the principle of conservation of momentum. The final momentum of the 0.32 kg ball is -0.153 kg*m/s.

Step-by-step explanation:

To solve this problem, we will use the principle of conservation of momentum. This principle states that the total momentum before a collision is equal to the total momentum after the collision, assuming there are no external forces acting on the system.

1. First, let's calculate the initial momentum of the system:

• The initial momentum of the 0.32 kg ball is given by: p1i = m1 * v1i = (0.32 kg) * (5.1 m/s)
• The initial momentum of the 0.75 kg ball is zero since it is initially at rest: p2i = 0

Next, let's calculate the final momentum of the system:

• The final momentum of the 0.32 kg ball is given by: p1f = m1 * v1f (which we need to find)
• The final momentum of the 0.75 kg ball is given by: p2f = m2 * v2f = (0.75 kg) * (2.38 m/s)

According to the principle of conservation of momentum, we have: p1i + p2i = p1f + p2fPlugging in the known values, we can solve for p1f:

p1f = p1i + p2i - p2f

p1f = (0.32 kg) * (5.1 m/s) + 0 - (0.75 kg) * (2.38 m/s)

p1f = (1.632 kg*m/s) - (1.785 kg*m/s)

p1f = -0.153 kg*m/s

Therefore, the final momentum of the 0.32 kg ball is -0.153 kg*m/s.