A boy of mass 52 kg jumps with a horizontal velocity of 2 m/s onto a stationary cart of mass 3 kg. The cart is fixed with frictionless wheels. Which one of the following would be the speed of the cart?

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Q: 31 (NDA-I/2022)
A boy of mass 52 kg jumps with a horizontal velocity of 2 m/s onto a stationary cart of mass 3 kg. The cart is fixed with frictionless wheels.
Which one of the following would be the speed of the cart?

question_subject: 

Science

question_exam: 

NDA-I

stats: 

0,17,7,4,17,2,1

keywords: 

{'horizontal velocity': [0, 0, 0, 1], 'stationary cart': [0, 0, 0, 1], 'speed': [0, 1, 2, 0], 'cart': [0, 0, 0, 1], 'frictionless wheels': [0, 0, 0, 1], 'kg': [0, 1, 9, 24], 'mass': [0, 0, 2, 3]}

Option 2 states that the speed of the cart would be 1-89 m/s.

Let`s analyze the given scenario. A boy with a mass of 52 kg jumps onto a stationary cart with a mass of 3 kg. The horizontal velocity of the boy is given as 2 m/s. Since there are no external horizontal forces acting on the system (no friction or external forces mentioned), the total momentum of the system before and after the jump must be conserved.

Before the jump, the momentum of the boy is given by mass x velocity, which is equal to 52 kg x 2 m/s = 104 kg*m/s. Since the cart is stationary initially, its momentum is zero.

After the jump, the boy and the cart together move with a common horizontal velocity. Let`s assume the speed of the cart after the jump is v m/s.

Using the principle of conservation of momentum, we can set up the equation: momentum before = momentum after.

104 kg*m/s + 0 kg*m/s = (52 kg + 3 kg) x v

Simplifying this equation, we get:

104 kg*m/s = 55 kg x v

v = 104 kg*m/s / 55 kg = 1.

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