Question

The change in kinetic energy of the system when the man jumps off the trolley is

• A. 1080 J
• B. 980 J
• C. 1180 J
• D. 1280 J

Answer 1

Answer: (A)

1080 J

Kinetic Energy $=\frac{1}{2}M{V}^2$

At initial state: $K.E.=\frac{1}{2}(80+40)(2{)}^2=240J$

From conservation of momentum, we have:

$(80+40)\left(2\right)=\left(80\right(-1)+40V_{trolley}$

$\Rightarrow V_{trolley}=8m/s$

At final state : $K.E.=\left[\frac{1}{2}{80.1}^2\right]+\left[\frac{1}{2}{40.8}^2\right]=1320J$

Change in $K.E.=1320-240=1080J$