A block of mass $$0.50\,kg$$ is moving with a speed of $$2.00\,m{s^{ - 1}}$$ on a smooth surface. It strikes another mass of $$1.00\,kg$$ and then they move together as a single body. The energy loss during the collision is
A.
$$0.16\,J$$
B.
$$1.00\,J$$
C.
$$0.67\,J$$
D.
$$0.34\,J$$
Answer :
$$0.67\,J$$
Solution :
Initial kinetic energy of the system
$$K.{E_i} = \frac{1}{2}m{u^2} + \frac{1}{2}M{\left( 0 \right)^2} = \frac{1}{2} \times 0.5 \times 2 \times 2 + 0 = 1\,J$$
For collision, applying conservation of linear momentum
$$\eqalign{
& m \times u = \left( {m + M} \right) \times v \cr
& \therefore 0.5 \times 2 = \left( {0.5 + 1} \right) \times v \Rightarrow v = \frac{2}{3}\,m/s \cr} $$
Final kinetic energy of the system is
$$K.{E_f} = \frac{1}{2}\left( {m + M} \right){v^2} = \frac{1}{2}\left( {0.5 + 1} \right) \times \frac{2}{3} \times \frac{2}{3} = \frac{1}{3}\,J$$
$$\therefore $$ Energy loss during collision $$ = \left( {1 - \frac{1}{3}} \right)J = 0.67\,J$$
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-