The coefficient of restitution $$e$$ for a perfectly elastic collision is
A.
$$1$$
B.
zero
C.
infinite
D.
$$- 1$$
Answer :
$$1$$
Solution :
The degree of elasticity of a collision is determined by a quantity called coefficient of restitution or coefficient of resilience of the collision. It is defined as the ratio of relative velocity of separation after collision to the relative velocity of approach before collision. It is represented by $$e.$$
$$\eqalign{
& e = \frac{{{\text{relative velocity of separation}}\left( {{\text{after collision}}} \right)}}{{{\text{relative velocity of approach}}\left( {{\text{before collision}}} \right)}} \cr
& e = \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}} \cr} $$
where, $${u_1},{u_2}$$ are the velocities of two bodies before collision and $${v_1},{v_2}$$ are their respective velocities after collision.
For a perfectly elastic collision, relative velocity of separation after collision is equal to relative velocity of approach before collision
$$\therefore e = 1$$
Releted MCQ Question on Basic Physics >> Work Energy and Power
Releted Question 1
If a machine is lubricated with oil-
A.
the mechanical advantage of the machine increases.
B.
the mechanical efficiency of the machine increases.
C.
both its mechanical advantage and efficiency increase.
D.
its efficiency increases, but its mechanical advantage decreases.
A particle of mass $$m$$ is moving in a circular path of constant radius $$r$$ such that its centripetal acceleration $${a_c}$$ is varying with time $$t$$ as $${a_c} = {k^2}r{t^2}$$ where $$k$$ is a constant. The power delivered to the particles by the force acting on it is:
A.
$$2\pi m{k^2}{r^2}t$$
B.
$$m{k^2}{r^2}t$$
C.
$$\frac{{\left( {m{k^4}{r^2}{t^5}} \right)}}{3}$$
A spring of force-constant $$k$$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force-constant of-