Question

A board is moving with velocity v on a rough horizontal plane. The upper surface of the board smooth on which a ball falls with velocity v and rebounds with velocity $\frac{v}{2}$. The mass of the board same as that of ball. After the collision, the board comes to state of rest. The co-efficient of friction between the board and the ground can be:

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{4}$
• D. $\frac{3}{5}$

Answer 1

Answer: (B)

$\frac{2}{3}$

Answer 2

The problem involves analyzing the impulse and momentum of a board and a ball during a collision. Here's a breakdown:

1. Horizontal Momentum Change of the Board:

   • The board's initial horizontal momentum is $Mv$, and it comes to rest, so the final momentum is 0.
   • The change in momentum, $-Mv$, is due to the impulse of kinetic friction from the ground.

2. Vertical Impulse on the Board:

   • The ball's vertical momentum changes from $-mv$ to $m(v/2)$.
   • This change, $\frac{3}{2}mv$, represents the vertical impulse exerted by the board on the ball.
   • By Newton's third law, the ball exerts an equal and opposite vertical impulse ($\frac{3}{2}Mv$, since $m=M$) downwards on the board.

3. Relating Friction Impulse to Normal Force:

   • The horizontal friction impulse is $J_x = \int \mu N(t) dt$.
   • The normal force $N(t)$ during the collision is $Mg$ (board's weight) plus the instantaneous vertical force from the ball $F_{ball,y}(t)$.

4. Setting up the Impulse-Momentum Equation:

   • $Mv = \mu (Mg\Delta t + \int F_{ball,y}(t) dt)$.
   • The integral $\int F_{ball,y}(t) dt$ is the vertical impulse $J_{by} = \frac{3}{2}Mv$.

5. Solving for $\mu$ (Coefficient of Friction):

   • $Mv = \mu (Mg\Delta t + \frac{3}{2}Mv)$.
   • Assuming the collision duration $\Delta t$ is negligible for the $Mg\Delta t$ term compared to the impulse from the ball, we get $Mv = \mu (\frac{3}{2}Mv)$, which simplifies to $\mu = \frac{2}{3}$.

Therefore, the coefficient of friction between the board and the ground can be $\frac{2}{3}$.