Question

a solid billiard ball with a mass of 230g and a radius of 4.00 cm is rolling across a flat surface with a speed of 1.35 m/s. it encounters an incline at a 20° with respect to the horizontal. how far up the incline does the ball roll before it stops? answer this question using the principle of energy conservation. use g = 10 m/s² for the acceleration due to gravity.

Explanation:

Step1: Recall Energy Conservation for Rolling Object

A rolling solid sphere (billiard ball is a solid sphere) has both translational and rotational kinetic energy. The moment of inertia \( I \) for a solid sphere is \( I=\frac{2}{5}mr^{2} \). The total initial kinetic energy \( K_{i} \) is the sum of translational \( K_{t}=\frac{1}{2}mv^{2} \) and rotational \( K_{r}=\frac{1}{2}I\omega^{2} \). Since \( v = r\omega \), \( \omega=\frac{v}{r} \). Substituting \( I \) and \( \omega \) into \( K_{r} \): \( K_{r}=\frac{1}{2}(\frac{2}{5}mr^{2})(\frac{v^{2}}{r^{2}})=\frac{1}{5}mv^{2} \). So total initial kinetic energy \( K_{i}=\frac{1}{2}mv^{2}+\frac{1}{5}mv^{2}=\frac{7}{10}mv^{2} \).

Step2: Final Energy (Potential Energy)

When the ball stops at height \( h \) up the incline, all kinetic energy is converted to gravitational potential energy \( U_{f}=mgh \). By energy conservation \( K_{i}=U_{f} \), so \( \frac{7}{10}mv^{2}=mgh \). The mass \( m \) cancels out, and we can solve for \( h \): \( h = \frac{7v^{2}}{10g} \).

Step3: Relate Height to Distance on Incline

Let the distance up the incline be \( d \). From trigonometry, \( \sin\theta=\frac{h}{d} \), so \( h = d\sin\theta \). Substitute \( h \) into the energy equation: \( d\sin\theta=\frac{7v^{2}}{10g} \), then \( d=\frac{7v^{2}}{10g\sin\theta} \).

Step4: Substitute Values

Given \( v = 1.35\ m/s \), \( g = 10\ m/s^{2} \), \( \theta = 20^{\circ} \), \( \sin20^{\circ}\approx0.3420 \). Plug in the values: \( d=\frac{7\times(1.35)^{2}}{10\times10\times0.3420} \). Calculate numerator: \( 7\times1.8225 = 12.7575 \). Denominator: \( 10\times10\times0.3420 = 34.2 \). Then \( d=\frac{12.7575}{34.2}\approx0.373\ m \)? Wait, no, wait, mistake in step 2: Wait, \( g = 10\ m/s^{2} \), so denominator is \( 10\times10\times0.3420 \)? No, wait, \( 10g\sin\theta \): \( g = 10 \), so \( 10\times10\times\sin20^{\circ} \)? No, wait the formula was \( d=\frac{7v^{2}}{10g\sin\theta} \). So \( 10g = 10\times10 = 100 \), times \( \sin20^{\circ}\approx0.3420 \), so denominator is \( 100\times0.3420 = 34.2 \). Numerator: \( 7\times(1.35)^2 = 7\times1.8225 = 12.7575 \). Then \( d=\frac{12.7575}{34.2}\approx0.373\ m \)? Wait, that seems low. Wait, maybe I messed up the moment of inertia? Wait, billiard ball is a solid sphere, \( I=\frac{2}{5}mr^{2} \), so rotational KE is \( \frac{1}{2}I\omega^{2}=\frac{1}{2}\times\frac{2}{5}mr^{2}\times\frac{v^{2}}{r^{2}}=\frac{1}{5}mv^{2} \), translational KE is \( \frac{1}{2}mv^{2} \), total KE \( \frac{7}{10}mv^{2} \), that's correct. Then \( h=\frac{7v^{2}}{10g}=\frac{7\times1.8225}{100}=\frac{12.7575}{100}=0.127575\ m \). Then \( d = \frac{h}{\sin\theta}=\frac{0.127575}{0.3420}\approx0.373\ m \)? Wait, no, wait \( h = d\sin\theta \), so \( d = h / \sin\theta \), and \( h = 7v²/(10g) \), so \( d = (7v²)/(10g\sin\theta) \). Let's recalculate: \( v = 1.35 \), \( v² = 1.8225 \), \( 7\times1.8225 = 12.7575 \), \( 10g = 100 \), \( \sin20°≈0.3420 \), so \( 10g\sin\theta = 100\times0.3420 = 34.2 \), then \( d = 12.7575 / 34.2 ≈ 0.373\ m \). Wait, but let's check units: \( v \) in m/s, \( g \) in m/s², so \( v²/g \) is in m, times 7/10, then divided by sinθ, so units are m, correct. Alternatively, maybe I made a mistake in moment of inertia? Wait, solid sphere: yes, \( I = 2/5 mr² \), so rotational KE is 1/5 mv², translational 1/2 mv², total 7/10 mv², correct. So the calculation seems right.

Wait, wait, no! Wait, the problem says "solid billiard ball" – is a billiard ball a solid sphere? Yes, typically. So the energy conservation should include both translational and rotational KE. So the steps are correct. Let's recalculate with more precise sin20°: sin20° ≈ 0.3420201433. So \( d = (7*(1.35)^2)/(10*10*0.3420201433) \). \( 1.35^2 = 1.8225 \), 7*1.8225=12.7575. Denominator: 10*10=100, 100*0.3420201433=34.20201433. Then 12.7575 / 34.20201433 ≈ 0.373 meters, which is 37.3 cm. That seems reasonable? Wait, let's check with just translational KE (if it were sliding, not rolling). Then KE would be 0.5 mv², so h = 0.5 v² / g = 0.5*1.8225 /10 = 0.091125 m, then d = 0.091125 / 0.3420 ≈ 0.266 m. But since it's rolling, it has more initial KE (rotational), so it should go further, which 0.373 m is further than 0.266 m, so that makes sense. So the calculation is correct.

Answer:

The distance up the incline is approximately \(\boxed{0.373\ m}\) (or more precisely, let's recalculate with exact steps:

Wait, wait, I think I messed up the formula in step 2. Wait, total initial KE is \( \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \). For solid sphere, \( I = \frac{2}{5}mr^2 \), \( \omega = v/r \), so \( \frac{1}{2}I\omega^2 = \frac{1}{2} \times \frac{2}{5}mr^2 \times \frac{v^2}{r^2} = \frac{1}{5}mv^2 \). So total KE is \( \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{5}{10}mv^2 + \frac{2}{10}mv^2 = \frac{7}{10}mv^2 \), correct. Then set equal to \( mgh \), so \( \frac{7}{10}v^2 = gh \), so \( h = \frac{7v^2}{10g} \). Then \( h = d\sin\theta \), so \( d = \frac{7v^2}{10g\sin\theta} \). Plugging in \( v = 1.35 \), \( g = 10 \), \( \sin20° \approx 0.3420 \):

\( d = \frac{7 \times (1.35)^2}{10 \times 10 \times 0.3420} = \frac{7 \times 1.8225}{34.2} = \frac{12.7575}{34.2} \approx 0.373 \) meters. So the answer is approximately 0.37 meters (or 37 cm). So \(\boxed{0.37\ m}\) (or more accurately 0.373 m).