Question

A billiard ball initially at rest is given a sharp blow by a cue stick. The impulse is horizontal and is applied at a distance 2R/3 below the centreline, as shown in figure. The initial speed of the ball is $v_0$, and the coefficient of kinetic friction is $\mu_k$.

• A. Initially kinetic friction acts in -$\hat{i}$ direction
• B. Initially kinetic friction acts in $\hat{i}$ direction.
• C. Ball instantaneously starts pure rolling.
• D. Initial angular velocity of ball is $\frac{5v_0}{3R}$

Answer 1

Answer: (B), (D)

B,D

Answer 2

1. Linear Motion Analysis:

The impulse $J$ is applied horizontally in the positive x-direction.

According to the impulse-momentum theorem, the change in linear momentum of the center of mass (CM) is equal to the impulse.

• Initial linear momentum $P_{initial} = 0$.
• Final linear momentum $P_{final} = Mv_{cm}$, where $M$ is the mass of the ball and $v_{cm}$ is the velocity of its center of mass.

Thus, $J = Mv_{cm}$.

Given that the initial speed of the ball is $v_0$ (meaning the speed of the center of mass immediately after the blow), we have $v_{cm} = v_0$ in the positive x-direction.

So, $J = Mv_0$.

2. Rotational Motion Analysis:

The impulse is applied at a distance $h = 2R/3$ below the centreline. This point is $R - 2R/3 = R/3$ from the bottom, or $2R/3$ below the center.

This impulse creates a torque about the center of mass. The magnitude of the torque is $\tau = J \times h = J \times (2R/3)$.

The direction of this torque is clockwise.

According to the angular impulse-angular momentum theorem, the change in angular momentum about the center of mass is equal to the angular impulse.

• Initial angular momentum $L_{initial} = 0$.
• Final angular momentum $L_{final} = I\omega_0$, where $I$ is the moment of inertia of the ball (a solid sphere) about its center, and $\omega_0$ is its initial angular velocity.

For a solid sphere, the moment of inertia is $I = \frac{2}{5}MR^2$.

The angular impulse is $J \times (2R/3)$.

Therefore, we have:

$I\omega_0 = J \times \frac{2R}{3}$

Substitute $I = \frac{2}{5}MR^2$ and $J = Mv_0$:

$\frac{2}{5}MR^2 \omega_0 = Mv_0 \left(\frac{2R}{3}\right)$

Cancel $M$ and $R$ from both sides:

$\frac{2}{5}R \omega_0 = \frac{2}{3}v_0$

Solve for $\omega_0$:

$\omega_0 = \frac{2}{3}v_0 \times \frac{5}{2R} = \frac{5v_0}{3R}$

The direction of $\omega_0$ is clockwise.

3. Velocity of the Point of Contact:

The velocity of the bottommost point (point of contact with the ground) is given by $v_P = v_{cm} - \omega R$. (Here, $v_{cm}$ is in the positive x-direction, and $\omega$ is clockwise, so its contribution to the tangential velocity at the bottom is in the negative x-direction, hence $-\omega R$).

Substitute the values of $v_{cm}$ and $\omega_0$:

$v_P = v_0 - \left(\frac{5v_0}{3R}\right)R$

$v_P = v_0 - \frac{5v_0}{3}$

$v_P = \frac{3v_0 - 5v_0}{3} = -\frac{2v_0}{3}$

4. Evaluate the Options:

• (A) Initially kinetic friction acts in -$\hat{i}$ direction:

  Since $v_P = -\frac{2v_0}{3}$, the point of contact is sliding to the left (in the $-\hat{i}$ direction) relative to the ground. Kinetic friction opposes relative motion, so it will act on the ball in the opposite direction, i.e., in the $+\hat{i}$ direction. Thus, (A) is incorrect.

• (B) Initially kinetic friction acts in $\hat{i}$ direction:

  As explained above, since the point of contact slides in the $-\hat{i}$ direction, friction acts on the ball in the $+\hat{i}$ direction. Thus, (B) is correct.

• (C) Ball instantaneously starts pure rolling:

  For pure rolling, the velocity of the point of contact $v_P$ must be zero. Since $v_P = -\frac{2v_0}{3} \neq 0$, the ball does not instantaneously start pure rolling. Thus, (C) is incorrect.

• (D) Initial angular velocity of ball is $\frac{5v_0}{3R}$:

  From our calculation in step 2, the initial angular velocity $\omega_0 = \frac{5v_0}{3R}$. Thus, (D) is correct.