When tossed upward and hit horizontally by a batter, a 0.23 kg softball receives an impulse of 2.0 N·s. With what horizontal speed does the ball move away from the bat?

To determine the horizontal speed at which the softball moves away from the bat, we can use the concept of impulse-momentum.

The formula for impulse is:
Impulse = Force × Time

Knowing that the impulse received by the softball is 2.0 N·s, we can rearrange the formula to solve for force:

Force = Impulse / Time

Since no time value is given in the problem, we can assume that the contact time between the bat and the ball is very short, which means the force can be considered constant during this interval.

Now, we need to calculate the horizontal component of the force. Here, the vertical component of the force will not contribute to the ball's horizontal speed.

Impulse = Change in momentum

Since the softball is tossed upward, the vertical component of momentum will be zero initially and finally. Thus, the impulse acting on the horizontal component of the momentum can be considered equal to the impulse given in the problem.

Now, we know the relationship between impulse and change in momentum:

Impulse = Change in momentum

Substituting the values into the equation:
2.0 N·s = Change in momentum

Next, we need to calculate the change in momentum. As we mentioned earlier, the vertical component of momentum doesn't change, so the change in momentum is only in the horizontal direction.

Change in momentum = Final momentum - Initial momentum

Since the ball was initially at rest, the initial momentum is zero:
Initial momentum = 0

The final momentum is equal to the product of mass and velocity:
Final momentum = mass × velocity

Since the mass is given as 0.23 kg, we can rewrite the equation as follows:
2.0 N·s = 0.23 kg × velocity

Now, solving for velocity:
velocity = (2.0 N·s) / (0.23 kg)

Calculating the value:
velocity ≈ 8.7 m/s

Therefore, the softball moves away from the bat with a horizontal speed of approximately 8.7 m/s.

To find the horizontal speed with which the ball moves away from the bat, we need to use the impulse-momentum principle. Impulse is defined as the change in momentum of an object, and momentum is the product of mass and velocity.

The impulse is given as 2.0 N·s, and the mass of the softball is 0.23 kg. We need to find the horizontal velocity (speed) of the ball.

The impulse-momentum principle states that the change in momentum is equal to the impulse applied to an object. Mathematically, we can express it as:

Impulse = Change in momentum

where Impulse = Force × time = (mass × acceleration) × time = mass × (final velocity - initial velocity)

Since the ball is hit horizontally, the vertical component of the velocity is not affected. Therefore, we only need to consider the horizontal component.

Let's assume the initial velocity of the softball is zero since it's tossed upwards, and we're interested in the horizontal speed when it moves away from the bat. So the equation becomes:

2.0 N·s = 0.23 kg × final velocity

Now, we can solve for the final velocity:

final velocity = 2.0 N·s / 0.23 kg

final velocity ≈ 8.70 m/s

Therefore, the softball moves away from the bat with a horizontal speed of approximately 8.70 m/s.

Impulse = (momentum change)

Use that equation to solve for the final speed.

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