Speed of a ball after collision

[Naldo6]

1. Homework Statement [/b]



A ball and bat, approaching one another each with the same speed of 1.9 m/s, collide. Find the speed of the ball after the collision. (Assume the mass of the bat is much much larger than the mass of the ball and that this is an elastic collision with no rotational motion).

2. Find the factor by which the KE of the ball increases due to the collision.

Could anyone help me?...

 

[tiny-tim]

A ball and bat, approaching one another each with the same speed of 1.9 m/s, collide. Find the speed of the ball after the collision. (Assume the mass of the bat is much much larger than the mass of the ball and that this is an elastic collision with no rotational motion).

2. Find the factor by which the KE of the ball increases due to the collision.

Hi Naldo6!

Hint: call the mass of the ball m, and the mass of the bat M, solve in the usual way, and then let M/m tend to infinity.

 

[baba89]

I would approach this problem by using the principles of conservation of momentum and conservation of kinetic energy. In an elastic collision, these two principles state that the total momentum and total kinetic energy of the system before and after the collision remain constant.

To find the speed of the ball after the collision, we can use the equation for conservation of momentum:

m1v1 + m2v2 = m1v1' + m2v2'

Where m1 and v1 represent the mass and initial velocity of the ball, m2 and v2 represent the mass and initial velocity of the bat, and v1' and v2' represent the final velocities of the ball and bat after the collision. Since the mass of the bat is much larger than the mass of the ball, we can assume that m2 is essentially infinite, making the equation simplify to:

m1v1 + m2v2 = m1v1'

Since the initial velocities of the ball and bat are equal (1.9 m/s), we can substitute that value in the equation and solve for the final velocity of the ball:

m1(1.9 m/s) = m1v1'

v1' = 1.9 m/s

Therefore, the speed of the ball after the collision is also 1.9 m/s.

To find the factor by which the kinetic energy of the ball increases due to the collision, we can use the equation for conservation of kinetic energy:

1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2

Again, since the initial velocities of the ball and bat are equal and the mass of the bat is much larger, we can simplify the equation to:

1/2m1v1^2 = 1/2m1v1'^2

v1'^2 = v1^2

v1' = v1

Therefore, the kinetic energy of the ball remains the same before and after the collision. The factor by which the KE increases is 1, meaning there is no change in KE.

In summary, the speed of the ball after the collision remains the same at 1.9 m/s and there is no change in kinetic energy. This is consistent with the principles of conservation of momentum and