Question

A train is running at a speed of 108 km/hr.An inspection cart is also moving with a speed 10m/s in the same direction as train. The train hits the cart $(e=\frac{1}{2}).$ The velocity of the cart after the collision is nearly

• A. $20m/s$
• B. $40m/s$
• C. $50m/s$
• D. cannot determine as the masses of train and cart are not given.

Answer 1

Answer: (B)

$40m/s$

A train is running at a speed of.....

$108km/h=108\times \frac{5}{18}=30m/s.$

From momentum conservation

$M\times 30+m\times 10=M{v}_1+m{v}_2....\left(i\right)$

e (velocity of approach) $=$ velocity of separation

$\Rightarrow \frac{1}{2}\times (30-10)=v_2-v_1$

$\Rightarrow 10=v_2-v_1....\left(ii\right)$
On solving,

$v_2=\frac{M\times 40+m\times 10}{M+m}m/s=\frac{40+m.\frac{10}{M}}{1+\frac{m}{M}}\approx 40m/s$