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:
- (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
