A man is standing on a cart of mass double the mass of the man. Initially cart is at rest on the smooth ground. Now man jumps with relative velocity ' v ' horizontally towards right with respect to cart. The work done by man during the process of jumping is

Question

A man is standing on a cart of mass double the mass of the man. Initially cart is at rest on the smooth ground. Now man jumps with relative velocity ' v ' horizontally towards right with respect to cart. The work done by man during the process of jumping is (A) m v2 (B) (m v2/2) (C) (m v2/4) (D) (m v2/3)

Tardigrade Admin · Accepted Answer

Let the velocity of man after jumping be '

u

' towards right.
Then speed of cart is

v - u

towards left.
From conservation of momentum

m u = 2 m (v - u)

∴ u = \frac{2 v}{3}

Hence, work done by man = change in

K E

of system

= \frac{1}{2} m u^{2} + \frac{1}{2} 2 m (v - u)^{2}

= \frac{1}{2} m (\frac{2 v}{3})^{2} + \frac{1}{2} 2 m (\frac{v}{3})^{2} = \frac{m v ^{2}}{3}

Q. A man is standing on a cart of mass double the mass of the man. Initially cart is at rest on the smooth ground. Now man jumps with relative velocity ' $v$ ' horizontally towards right with respect to cart. The work done by man during the process of jumping is

$m v^{2}$

$\frac{m v ^{2}}{2}$

$\frac{m v ^{2}}{4}$

$\frac{m v ^{2}}{3}$

Q. A man is standing on a cart of mass double the mass of the man. Initially cart is at rest on the smooth ground. Now man jumps with relative velocity ' v ' horizontally towards right with respect to cart. The work done by man during the process of jumping is

mv2

2mv2​

4mv2​

3mv2​

Let the velocity of man after jumping be ' u ' towards right. Then speed of cart is v−u towards left. From conservation of momentum mu=2m(v−u) ∴u=32v​ Hence, work done by man = change in KE of system =21​mu2+21​2m(v−u)2 =21​m(32v​)2+21​2m(3v​)2=3mv2​

Q. A man is standing on a cart of mass double the mass of the man. Initially cart is at rest on the smooth ground. Now man jumps with relative velocity ' $v$ ' horizontally towards right with respect to cart. The work done by man during the process of jumping is

$m v^{2}$

$\frac{m v ^{2}}{2}$

$\frac{m v ^{2}}{4}$

$\frac{m v ^{2}}{3}$