Question

After falling from rest through a height $h$, a body of mass $m$ begins to raise a body of mass $M(M>m)$ connected to it through a pulley. Find the fraction of kinetic energy lost when the body of mass $M$ is jerked into motion .

• A. $\frac{M}{M+m}$
• B. $\frac{M}{M-m}$
• C. $\frac{2M}{M+m}$
• D. $\frac{M}{2(M+m)}$

Answer 1

The correct option is A $\frac{M}{M+m}$

Initial energy $=K_1=\frac{1}{2}m{v}^2$

Initial momentum$=p_1=mv$

When M is jerked to motion, final velocity of both the mass is same say V as both are connected to a string.Let J be the impulse.

$MV=J$ and $mv-mV=J$

On solving-

$V=\frac{mv}{M+m}$

Final kinetic energy $=K_2=\frac{1}{2}(m+M)V^2=\frac{m^2{v}^2}{2(m+M)}$

This loss in kinetic energy =$E=K_1-K_2$

$E=K_1-K_2=\frac{mM{v}^2}{2(m+M)}$

Hence, required ratio$=\frac{E}{K_1}=\frac{M}{m+M}$

Answer-(A)