Question

A boy of mass 40 kg jumps with a horizontal velocity of 5 ms-1 onto a stationary cart with frictionless wheels. The mass of the cart is 3 kg. What is his velocity as the cart starts moving? Assume that there is no external unbalanced force working in horizontal direction​

Answer 1

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It is given that :-

• Mass of boy = 40kg
• Initial velocity of boy = 5m/s
• Mass of cart = 3kg
• Initial velocity of cart = 0(since it's at rest)
• Final velocity of cart = v = ?

Now according to conservation of linear momentum ,two separate bodies move together with same velocity after meeting. It can be represented by :

m1u1 + m2u2 = v(m1+m2)

After inputting the known values for this case we get : -

$\mapsto \: 40 \times 5 + 3 \times 0 = v(40 + 3)$

$\mapsto \: 200 = 43v$

$\mapsto \: v = \frac{200}{43}$

$\mapsto \: v = \huge\cancel \frac{200}{43}$

$\mapsto \: v = 4.65 \frac{m}{s}$

Therefore , The velocity when cart starts moving is 4.65m/s

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BY SCIVIBHANSHU

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