# Bat Hitting Ball: Find Speed After Collision

• alfredo24pr
In summary, the ball and bat collide with equal speeds of 2.30 m/s. This is an elastic collision with no rotational motion. By using the conservation of momentum and energy equations, and assuming the mass of the bat is much larger than the ball, the final velocity of the ball can be determined.
alfredo24pr

## Homework Statement

A ball and bat, approach one another each with the same speed of 2.30 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).

## Homework Equations

m1v1i + m2v2i = m1v1f + m2v2f

## The Attempt at a Solution

Im stuck at beggining :/

hi alfredo24pr!

elastic collision means that you use conservation of energy also

tiny-tim said:
hi alfredo24pr!

elastic collision means that you use conservation of energy also

hi tiny tim..so that means I have to relate kinetic energy?

(just got up :zzz: …)
alfredo24pr said:
hi tiny tim..so that means I have to relate kinetic energy?

yup!

what do you get?

tiny-tim said:
(just got up :zzz: …)

yup!

what do you get?

umm.. I do not understand. I know that I have to relate to v used in momentum to the v in kinetic energy

write out the conservation of momentum equation and the conservation of energy equation

tiny-tim said:
write out the conservation of momentum equation and the conservation of energy equation

m1v1i + m2v2i = m1v1f + m2v2f

and

1/2mvi2 + mghi = 1/2mvf2 + mghf

hi alfredo24pr!

ok, now use 2.30, and assume that m2/m1 is very much less than 1

tiny-tim said:
hi alfredo24pr!

ok, now use 2.30, and assume that m2/m1 is very much less than 1

so m2 / m1 is negligible therefore I can write it out as 0?

I still do not understand how to work with both equations.

start

tiny-tim said:
start

uhh..

m1v1 = m1(2.3)
m2v2 = m2(2.3)

1/2mvi2 = 1/2m(2.3)2

and then i substitute

m1(2.3) + m2(2.3) = m1vif + m2vif

1/2mvi2 + 0 = 1/2mvf2??

alfredo24pr said:
1/2mvi2 + 0 = 1/2mvf2??

what is this supposed to be?

(in particular, what is the "0" ?)

## 1. How does the speed of the bat affect the speed of the ball after collision?

The speed of the bat directly affects the speed of the ball after collision. This is because the bat is the object that is transferring its kinetic energy to the ball. The greater the speed of the bat, the greater the speed of the ball after collision.

## 2. What other factors besides speed of the bat can impact the speed of the ball after collision?

The mass of the bat and the mass of the ball can also impact the speed of the ball after collision. Additionally, the angle at which the bat strikes the ball and the elasticity of the ball can also play a role in determining the speed of the ball after collision.

## 3. How can we measure the speed of the ball after it is hit by a bat?

The speed of the ball after collision can be measured using a radar gun or a high-speed camera. These tools can accurately measure the speed of the ball in miles per hour (mph) or meters per second (m/s).

## 4. Is the speed of the ball after collision always the same as the speed of the bat?

No, the speed of the ball after collision is not always the same as the speed of the bat. This is because some of the kinetic energy from the bat is lost during the collision due to factors such as friction and air resistance.

## 5. Can we calculate the speed of the ball after collision using a formula?

Yes, the speed of the ball after collision can be calculated using the formula: v(ball) = (m(bat) * v(bat) + m(ball) * v(ball))/ (m(bat) + m(ball)), where v(ball) is the speed of the ball after collision, m(bat) is the mass of the bat, v(bat) is the speed of the bat, and m(ball) is the mass of the ball.

Replies
10
Views
2K
Replies
10
Views
2K
Replies
13
Views
477
Replies
2
Views
2K
Replies
4
Views
1K
Replies
20
Views
2K
Replies
34
Views
1K
Replies
4
Views
3K
Replies
1
Views
1K
Replies
18
Views
3K