Centre of Mass 4 Question 9
10. A wedge of mass $m$ and triangular cross-section $(A B=B C=C A=2 R)$ is moving with a constant velocity $(-v \hat{\mathbf{i}})$ towards a sphere of radius $R$ fixed on a smooth horizontal table as shown in the figure.
The wedge makes an elastic collision with the fixed sphere and returns along the same path without any rotation. Neglect all friction and suppose that the wedge remains in contact with the sphere for a very short time $\Delta t$ during which the sphere exerts a constant force $\mathbf{F}$ on the wedge.
$(1998,8$ M)
(a) Find the force $\mathbf{F}$ and also the normal force $\mathbf{N}$ exerted by the table on the wedge during the time $\Delta t$.
(b) Let $h$ denote the perpendicular distance between the centre of mass of the wedge and the line of action of $F$. Find the magnitude of the torque due to the normal force $\mathbf{N}$ about the centre of the wedge during the interval $\Delta t$.
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Answer:
Correct Answer: 10. (a) $\frac{2 m v}{\sqrt{3} \Delta t}(\sqrt{3} \hat{\mathbf{i}}-\hat{\mathbf{k}}), \frac{2 m v}{\sqrt{3} \Delta t}+m g \quad \hat{\mathbf{k}}$ (b) $\frac{4 m v}{\sqrt{3} \Delta t} h$
Solution:
- (a) (i) Since, the collision is elastic, the wedge will return with velocity $v \hat{\mathbf{i}}$.
Now, linear impulse in $x$-direction
$=$ change in momentum in $x$-direction.
$\therefore\left(F \cos 30^{\circ}\right) \Delta t=m v-(-m v)=2 m v$
$\therefore \quad F=\frac{2 m v}{\Delta t \cos 30^{\circ}}=\frac{4 m v}{\sqrt{3} \Delta t} \Rightarrow F=\frac{4 m v}{\sqrt{3} \Delta t}$
$\therefore \quad \mathbf{F}=\left(F \cos 30^{\circ}\right) \hat{\mathbf{i}}-\left(F \sin 30^{\circ}\right) \hat{\mathbf{k}}$
or $\quad \mathbf{F}=\frac{2 m v}{\Delta t} \hat{\mathbf{i}}-\frac{2 m v}{\sqrt{3} \Delta t} \hat{\mathbf{k}}$
(ii) Taking the equilibrium of wedge in vertical $z$-direction during collision.
$$ N=m g+F \sin 30^{\circ} \Rightarrow N=m g+\frac{2 m v}{\sqrt{3} \Delta t} $$
or in vector form
$$ \mathbf{N}=m g+\frac{2 m v}{\sqrt{3} \Delta t} \hat{\mathbf{k}} $$
(b) For rotational equilibrium of wedge [about (CM)] anticlockwise torque of $F=$ clockwise torque due to $N$
$\therefore$ Magnitude of torque of $N$ about $CM=$ magnitude of torque of $F$ about $CM$
$$ =F \cdot h\left|\tau _N\right|=\frac{4 m v}{\sqrt{3} \Delta t} h $$
