Question

In a Little League baseball game, the 145 g ball enters the strike zone with a speed of 17.0 m/s . The batter hits the ball, and it leaves his bat with a speed of 25.0 m/s in exactly the opposite direction. What is the magnitude of the impulse delivered by the bat to the ball? If the bat is in contact with the ball for 1.0 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 6.09 kg*m/s. The magnitude of the average force exerted by the bat on the ball during a contact time of 1.0 ms is 6090 N.

Step-by-step explanation:

The question relates to the concept of impulse and average force in a collision scenario. Impulse is given by the change in momentum of the object, and it can also be calculated as the average force (F) multiplied by the time duration (t) of the impact.

Considering the baseball scenario where the ball has a mass of 0.145 kg, enters the strike zone at 17.0 m/s and leaves in the opposite direction at a speed of 25.0 m/s, we can calculate the magnitude of the impulse delivered by the bat as follows:

• Impulse (J) = Change in momentum = m(v_final - v_initial)
• J = 0.145 kg * (25.0 m/s - (-17.0 m/s))
• J = 0.145 kg * 42.0 m/s
• J = 6.09 kg*m/s (This is the magnitude of the impulse)

To find the magnitude of the average force exerted by the bat on the ball, we use the time duration of the contact:

• Average force (F) = Impulse (J) / Time duration (t)
• F = 6.09 kg*m/s / 0.001 s
• F = 6090 N

So, the magnitude of the average force exerted by the bat on the ball is 6090 N.

Answer 2

impulse = 6.09 kg m/s

Force = 6090 N

Step-by-step explanation:

As we know that the impulse is defined as the change in momentum of the ball

so here we will have

`\Delta P = mv_f - mv_i`

now we know that

`v_f = 25 m/s`

initial speed is given as

`v_i = -17 m/s`

now impulse is given as

`\Delta P = 0.145(25 - (-17))`

`\Delta P = 6.09 kg m/s`

Now we also know that average force is defined as the rate of change in momentum

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

so we have

`F = (6.09)/(1 * 10^(-3))`

`F = 6.09 * 10^3 N`