A man and a cart move towards each other. The man weighs 64 kg and the cart 32 kg. The velocity of the man is 5.4 km/hr and that of the cart is 1.8km/hr. When the man approaches the cart, he jumps on to it. The velocity of the cart carrying the man will be:

30 km/hr

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3 km/hr

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1.8 km/ hr

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zero

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Solution

The correct option is C 3 km/hr
Using conservation of linear momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
Thus we get
$64 \times 1.5 - 32 \times 0.5 = 96 \times v$
$\Rightarrow v = \frac{80}{96} m / s$
$= 3 k m / h r$ in the direction of man.

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A man and a cart move towards each other. The man weighs 64 kg and the cart 32 kg. The velocity of the man is 5.4 km/hr and that of the cart is 1.8km/hr. When the man approaches the cart, he jumps on to it. The velocity of the cart carrying the man will be:

The correct option is C 3 km/hrUsing conservation of linear momentumm1u1+m2u2=m1v1+m2v2Thus we get64×1.5−32×0.5=96×v⇒v=8096m/s=3km/hr in the direction of man.

The correct option is C 3 km/hr
Using conservation of linear momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
Thus we get
$64 \times 1.5 - 32 \times 0.5 = 96 \times v$
$\Rightarrow v = \frac{80}{96} m / s$
$= 3 k m / h r$ in the direction of man.