Question

a bowling ball rolls down the lane at 4 m/s. the ball collides with the wall with a force of 64 n before coming to a complete stop 0.25 s later. what is the mass of the ball?

Explanation:

Step1: Recall Impulse-Momentum Theorem

The impulse-momentum theorem states that the impulse (\(J\)) applied to an object is equal to the change in its momentum (\(\Delta p\)). Mathematically, \(J = \Delta p\). Impulse is also given by \(J = F \cdot \Delta t\), and momentum is \(p = m \cdot v\), so the change in momentum is \(\Delta p = m \cdot \Delta v\).

Step2: Identify Known Values

• Initial velocity, \(v_i = 4 \, \text{m/s}\)
• Final velocity, \(v_f = 0 \, \text{m/s}\) (since it comes to a stop)
• Force, \(F = 64 \, \text{N}\)
• Time interval, \(\Delta t = 0.25 \, \text{s}\)

Step3: Calculate Change in Velocity

\(\Delta v = v_f - v_i = 0 - 4 = -4 \, \text{m/s}\) (the negative sign indicates a decrease in velocity)

Step4: Apply Impulse-Momentum Theorem

From \(J = \Delta p\), we have \(F \cdot \Delta t = m \cdot \Delta v\). We need to solve for mass \(m\), so rearranging the formula gives \(m = \frac{F \cdot \Delta t}{\Delta v}\).

Step5: Substitute Values and Solve

Substitute \(F = 64 \, \text{N}\), \(\Delta t = 0.25 \, \text{s}\), and \(\Delta v = -4 \, \text{m/s}\) into the formula:

$$ m = \frac{64 \cdot 0.25}{-4} $$

First, calculate the numerator: \(64 \cdot 0.25 = 16\)
Then, divide by \(-4\): \(m = \frac{16}{-4} = -4\). But mass can't be negative, so we take the magnitude (the negative sign indicates direction of force, not mass). So \(m = 4 \, \text{kg}\). Wait, no, wait. Wait, the force is applied by the wall, so the impulse should be in the opposite direction of the initial velocity. Let's recheck.

Wait, actually, the force is \(F = 64 \, \text{N}\) (the magnitude), and the change in velocity is \(\Delta v = -4 \, \text{m/s}\). But impulse \(J = F \cdot \Delta t\) (taking magnitude for force, but actually, force is a vector. Let's do it with signs. Let the initial direction be positive. Then \(F = -64 \, \text{N}\) (since it's opposite to the motion), \(\Delta t = 0.25 \, \text{s}\), \(\Delta v = -4 \, \text{m/s}\). Then \(J = F \cdot \Delta t = -64 \cdot 0.25 = -16 \, \text{N·s}\). \(\Delta p = m \cdot \Delta v = m \cdot (-4)\). So \(-16 = -4m\), so \(m = \frac{16}{4} = 4 \, \text{kg}\). Wait, that's correct. So the mass is \(4 \, \text{kg}\)? Wait, no, wait, 64 * 0.25 is 16, divided by 4 (since |Δv| is 4) gives 4? Wait, no, 64 N force, time 0.25 s, change in velocity 4 m/s (from 4 to 0). So \(F \cdot t = m \cdot \Delta v\), so \(m = \frac{F \cdot t}{\Delta v}\). So \(F = 64\), \(t = 0.25\), \(\Delta v = 4\) (magnitude). So \(m = \frac{64 \times 0.25}{4} = \frac{16}{4} = 4 \, \text{kg}\). Yes, that's correct.

Answer:

The mass of the bowling ball is \(\boxed{4}\) kilograms.