Laws of Motion - Result Question 7
7. A $0.5 kg$ ball moving with speed of $12 m / s$ strikes a hard wall at an angle of $30^{\circ}$ with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 seconds, the average force acting on the wall is
[2006]
(a) $24 N$
(b) $12 N$
(c) $96 N$
(d) $48 N$
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Answer:
Correct Answer: 7. (a)
Solution:
- (a)
The magnetitude of both velocity $v_1$ and $v_2$ are equal. Components of velocity along the wall are equal and are in same direction.
But the component of velocity perpendicular to wall are equal but opposite in direction.
So, net change in momentum in perpendicular direction,
$\Delta P=m v_1 \sin 30^{\circ}-(-m v_2 \sin 30^{\circ})$
$\Rightarrow 2 m v \sin 30^{\circ} \quad[|v_1| v_2 \mid]$
Average force acting on the wall will be given by
$ \begin{aligned} \Rightarrow & F x t=2 m v \sin 30^{\circ} \\ \Rightarrow & F=\frac{2 m v \sin 30^{\circ}}{t} \\ \Rightarrow & F=\frac{2 \times 0.5^{2} \times 12 \times 1}{2 \times 0.25}=24 \\ & F=24 N \end{aligned} $
