Physics College

. A 0.140 kg baseball is pitched toward home plate at 30.0 m/s.
The batter hits the ball back (opposite direction) to the pitcher at
44.0 m/s. Assume that towards home plate is positive. What is
the change in momentum for the ball?

Answers

Answer 1

The change in momentum for the baseball, which is hit back in the opposite direction by the batter, is -10.36 kg·m/s. This change in momentum is obtained by subtracting the initial momentum of 4.2 kg·m/s from the final momentum of -6.16 kg·m/s. The negative sign indicates the opposite direction of the momentum.

To find the change in momentum for the baseball, we can use the formula:

Change in momentum = Final momentum - Initial momentum

Momentum is defined as the product of mass and velocity.

Given data:

Mass of the baseball (m) = 0.140 kg

Initial velocity of the baseball ([tex]v_i_n_i_t_i_a_l)[/tex] = 30.0 m/s

Final velocity of the baseball ([tex]v_f_i_n_a_l_[/tex]) = -44.0 m/s (negative sign indicates opposite direction)

To calculate the initial momentum, we multiply the mass by the initial velocity:

Initial momentum = m * [tex]v_i_n_i_t_i_a_l[/tex] = 0.140 kg * 30.0 m/s = 4.2 kg·m/s

To calculate the final momentum, we multiply the mass by the final velocity:

Final momentum = m * [tex]v_f_i_n_a_l_[/tex] = 0.140 kg * (-44.0 m/s) = -6.16 kg·m/s

Now we can find the change in momentum:

Change in momentum = Final momentum - Initial momentum

Change in momentum = (-6.16 kg·m/s) - (4.2 kg·m/s)

Change in momentum = -10.36 kg·m/s

Therefore, the change in momentum for the baseball is -10.36 kg·m/s. The negative sign indicates a change in direction, as the ball is hit back in the opposite direction.

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Related Questions

Joe is painting the floor of his basement using a paint roller. The roller has a mass of 2.4 kg and a radius of 3.8 cm. In rolling the roller across the floor, Joe applies a force F = 16 N directed at an angle of 35° as shown. Ignoring the mass of the roller handle, what is the magnitude of the angular acceleration of the roller?

Answers

The magnitude of the angular acceleration of the roller is approximately 104.2 rad/s^2.

The magnitude of the angular acceleration of the roller can be determined using the torque equation and Newton's second law for rotational motion.
Step 1: Calculate the moment of inertia of the roller.
The moment of inertia (I) of a solid cylinder is given by the formula I = (1/2) * m * r^2, where m is the mass of the object and r is the radius.
In this case, the mass of the roller is 2.4 kg and the radius is 0.038 m.
So, I = (1/2) * 2.4 kg * (0.038 m)^2.
Step 2: Calculate the torque applied to the roller.
Torque (τ) is equal to the force (F) applied multiplied by the perpendicular distance (r) from the axis of rotation.
In this case, the force applied by Joe is 16 N and the distance is equal to the radius of the roller, 0.038 m.
So, τ = F * r.
Step 3: Use the torque equation.
The torque applied to the roller causes an angular acceleration (α) according to the equation τ = I * α.
Rearranging the equation, we get α = τ / I.
Step 4: Substitute the values into the equation.
Using the values we calculated earlier, we can substitute them into the equation α = τ / I.
α = (16 N * 0.038 m) / [(1/2) * 2.4 kg * (0.038 m)^2].
Step 5: Calculate the magnitude of the angular acceleration.
Evaluating the expression, we find that the magnitude of the angular acceleration of the roller is approximately 104.2 rad/s^2.
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