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?
[amp_mcq option1=”2.15 m/s” option2=”1.89 m/s” option3=”1.51 m/s” option4=”2.51 m/s” correct=”option2″]
This question was previously asked in
UPSC NDA-1 – 2022
– Before the jump, the boy has momentum (mass × velocity) and the cart has zero momentum.
– After the boy jumps onto the cart, they move together as a single system with a common velocity.
– According to the law of conservation of momentum, the total momentum of the system before the collision is equal to the total momentum after the collision, provided no external forces act on the system (frictionless wheels are mentioned).
– Initial momentum = (mass of boy × velocity of boy) + (mass of cart × velocity of cart) = (52 kg × 2 m/s) + (3 kg × 0 m/s) = 104 kg m/s.
– Let the final velocity of the combined system (boy + cart) be ‘v’. The combined mass is 52 kg + 3 kg = 55 kg.
– Final momentum = (combined mass × final velocity) = 55 kg × v.
– By conservation of momentum: 104 kg m/s = 55 kg × v.
– v = 104 / 55 ≈ 1.89 m/s.