Question

A ball of mass 1.2kg strike a wall at an angle of 30° with velocity 5m/s and rebound at an angle of 60 with a velocity of 2m/s. The angles are with respect to the line perpendicular to wall. What is the coefficient of friction between the ball and the wall. Assume the friction is kinr

• A. 0.144

Answer 1

Answer: (A)

0.144

Answer 2

Solution:

1. Set Up Coordinate System:
   Let the $y$-axis be normal (perpendicular) to the wall (positive outward from the wall) and the $x$-axis be tangential along the wall.

2. Resolve Velocities:

   • Before collision: $$u_y = -5\cos30^\circ = -5\left(\frac{\sqrt3}{2}\right) = -\frac{5\sqrt3}{2},\quad u_x = 5\sin30^\circ = 5\left(\frac{1}{2}\right) = 2.5$$
   • After collision: $$v_y = 2\cos60^\circ = 2\left(\frac{1}{2}\right) = 1,\quad v_x = 2\sin60^\circ = 2\left(\frac{\sqrt3}{2}\right) = \sqrt3$$

3. Impulses (Change in Momentum):
   With mass $m = 1.2\,\text{kg}$:

   • Normal (y-direction) Impulse: $$\Delta p_y = m(v_y - u_y) = 1.2\left(1 - \left(-\frac{5\sqrt3}{2}\right)\right) = 1.2\left(1 + \frac{5\sqrt3}{2}\right)$$
   • Tangential (x-direction) Impulse: $$\Delta p_x = m(v_x - u_x) = 1.2\left(\sqrt3 - 2.5\right)$$ (Here $\sqrt3$ is approximately 1.732, so the term in parentheses is negative indicating the reversal in the tangential effect.)

4. Coefficient of Kinetic Friction ($\mu$):
   The frictional impulse is responsible for the change in tangential momentum and is given by:

   $$\mu = \frac{|\Delta p_x|}{\Delta p_y} = \frac{1.2\left(2.5 - \sqrt3\right)}{1.2\left(1+\frac{5\sqrt3}{2}\right)} = \frac{2.5 - \sqrt3}{1+\frac{5\sqrt3}{2}}$$

5. Final Calculation:
   Substitute numerical approximations ($\sqrt3 \approx 1.732$):

   $$2.5 - \sqrt3 \approx 2.5 - 1.732 = 0.768, \quad 1+\frac{5\sqrt3}{2} \approx 1+\frac{5 \times 1.732}{2} = 1+4.33 = 5.33$$

   Thus,

   $$\mu \approx \frac{0.768}{5.33} \approx 0.144$$

Explanation (Minimal):

• Resolve the velocities into normal and tangential components using the given angles.
• Compute the changes in momentum in the normal and tangential directions.
• The coefficient of kinetic friction is given by the ratio of the magnitude of the tangential impulse to the normal impulse.
• Simplify to get $\mu \approx 0.144$.