Question

When a 0.15 kg baseball is hit, its velocity changes from +17 m/s to -17 m/s.

(a) What is the magnitude of the impulse delivered by the bat to the ball?
N·s

(b) If the baseball is in contact with the bat for 1.5 ms, what is the magnitude of the average force exerted by the bat on the ball?

Answer 1

The magnitude of the impulse delivered by the bat to the ball is 5.1 kg·m/s. The magnitude of the average force exerted by the bat on the ball is 3.4 × 10^3 N.

Step-by-step explanation:

To find the magnitude of the impulse delivered by the bat to the ball, we need to use the formula for impulse:

Impulse = change in momentum

Momentum is given by the product of mass and velocity (momentum = mass × velocity). The change in momentum is the final momentum minus the initial momentum (change in momentum = final momentum - initial momentum).

Given that the mass of the baseball is 0.15 kg, the initial velocity is +17 m/s, and the final velocity is -17 m/s, we can calculate the magnitude of the impulse:

Impulse = (0.15 kg × -17 m/s) - (0.15 kg × 17 m/s) = -5.1 kg·m/s

So, the magnitude of the impulse delivered by the bat to the ball is 5.1 kg·m/s.

To find the magnitude of the average force exerted by the bat on the ball, we can use the formula for average force:

Average force = impulse / time

Given that the time the ball is in contact with the bat is 1.5 ms (1.5 × 10^-3 s) and the magnitude of the impulse is 5.1 kg·m/s, we can calculate the magnitude of the average force:

Average force = 5.1 kg·m/s / (1.5 × 10^-3 s) = 3.4 × 10^3 N

Therefore, the magnitude of the average force exerted by the bat on the ball is 3.4 × 10^3 N.

Answer 2

Step-by-step explanation:

It is given that,

Mass of the baseball, m = 0.15 kg

Initial velocity, u = +17 m/s

Final velocity, v = -17 m/s

(a) We need to find the magnitude of impulse delivered by the bat to the ball. We know that impulse is equal to the change in momentum of an object i.e.

`J=\Delta p=m(v-u)`

`J=0.15\ kg* (-17\ m/s-17\ m/s)`

J = −5.1 Kg-m/s

(b) If the baseball is in contact with the bat for 1.5 ms i.e. the time of collision Δt = 0.0015 s

We need to find the magnitude of the average force exerted by the bat on the ball.

`J=F.\Delta t`

`F=(J)/(\Delta t)`

`F=(5.1\ Kg-m/s)/(0.0015\ s)`

F = 3400 N

Hence, this is the required solution.