Laws Of Motion Ques 46
- A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in
(2001, 2M)
(a)
(b)
(c)
(d)
Show Answer
Answer:
Correct Answer: 46.(a)
Solution:
- Since, the block rises to the same heights in all the four cases, from conservation of energy, speed of the block at highest point will be same in all four cases. Say it is $v_{0}$.
Equation of motion will be
$$ \begin{aligned} N+m g & =\frac{m v_{0}^{2}}{R} \\ or && N & =\frac{m v_{0}^{2}}{R}-m g \end{aligned} $$
$R$ (the radius of curvature) in first case is minimum. Therefore, normal reaction $N$ will be maximum in first case.
NOTE
In the question, it should be mentioned that all the four tracks are frictionless. Otherwise, $\mathrm{v}_{0}$ will be different in different tracks.
BETA
