Question

A girl of mass 40 kg jumps with a horizontal velocity of 5 m/s onto a stationary cart with frictionless wheels. The Mass of the cart is 3 kg. What is velocity as the cart starts moving? Assume that there is no external unbalanced force working in horizontal direction .

Answer 1

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Let v be the velocity of the girl on the cart as the cart starts moving
The total momentum of the girl and the cart before interaction
= m×v + m×v

= 40 × 5 + 3 × 0

= 200 kg m/s

The total momentum after the interaction

= ( 40 + 3 ) kg × v m/s

= 43v kg m/s


According to the law of conservation of momentum, the total momentum is conserved during the interaction. So,

》 43v = 200

》 V = $\frac{200}{43}$

》V = 4.65 m/s $\boxed{Ans}$


The girl on the cart would move with Velocity of 4.65 m/s in the direction in which the girl jump onto the cart.


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Hope this helps. . . . . . . . ☺☺

Answer 2

Mass of girl = m1 = 40 kg

Mass of cart = m2 = 3 kg

Initial velocity of the girl = u1 = 5m/s

Initial velocity of the cart = u2 = 0 m/s

Let the final velocity of the cart and the girl be v

∴ According to the law of conservation of momentum,

m1u1 + m2u2 = m1v1 + m2v2

⇒ (40 × 5) + (3 × 0) = (40 × v) + (3 × v)

⇒ 200 + 0 = 40 v + 3 v

⇒ 43 v = 200

⇒ v = 200/3 = 4.65 m/s

∴ The velocity of the cart after the girl starts moving will be 4.65 m/s .