Laws of Motion - Result Question 9

9. A3 kg ball strikes a heavyrigid wall with a speed of $10 m / s$ at an angle of $60^{\circ}$. It gets reflected with the same speed and angle as shown here. If the ball is in contact with the wall for $0.20 s$, what is the average forceexerted on the ball by the wall?

(a) $150 N$

(b) zero

(c) $150 \sqrt{3} N$

(d) $300 N$

[2000]

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Answer:

Correct Answer: 9. (c)

Solution:

  1. (c) Change in momentum along the wall $=m v \cos 60^{\circ}-m v \cos 60^{\circ}=0$

Change in momentum perpendicular to the wall $=m v \sin 60^{\circ}-(-m v \sin 60^{\circ})=2 m v \sin 60^{\circ}$

$\therefore$ Applied force $=\frac{\text{ Change in momentum }}{\text{ Time }}$

$=\frac{2 m v \sin 60^{\circ}}{0.20}$

$=\frac{2 \times 3 \times 10 \times \sqrt{3}}{2 \times 20}=50 \times 3 \sqrt{3}$

$=150 \sqrt{3}$ newton