Question

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A body of mass 40 kg jump with the horizontal velocity of 5 metre per second onto a stationary cart with frictionless wheel. The mass of cart is 3 kg , what is the velocity as the cart start moving.

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$\color{red}{Class - 9th}$

$\color{red}{Sub - Physics}$

$\color{red}{Answer: 4.65 (not\: sure)}$

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Answer 1

From conservation of liner momentum
MV = (M + m)v
v = MV/(M + m)
v = (40×5)/43
v = 200/43 = 4.65 m/s ( approx)

Answer 2

$Hey \: !!$

Here is your answer...

Given,

$mass \: of \: body = 40 \: kg$

$velocity = 5 \: m /s$

$mass \: of \: cart = 3 \: kg$

Since, no external force is there, therefore the momentum is constant.

Now,

$momentum \: of \: girl = 40 \times 5 = 200 \: kg.m/s$

Since, momentum of cart is zero therfore total momentum will be equal to momentum of girl i.e., 200 kg.m/s .

Also,

$velocity= 200 \div (40 + 3) = 4.65 \: m/s$

So the velocity will be 4.65 m/s .

Hope it helps...