A girl of mass 40 kg jumps with a horizontal velocity of 5 m/s onto a cart of mass
10kg with friction less wheels, which is already moving with a horizontal velocity 1 m/s
in the opposite direction of motion of the girl. What will be the velocity of the girl, when
the cart moves along with her. (Assume that there is no external unbalanced force
working in the horizontal direction)
Answers
Answer:
:law of conservation of momentum. mass of girl(m1) = 40 kg initial velocity of girl(u1) = 5m/s mass of cart(m2) = 10 kg initial vel. of cart(u2) = 1m/s final velocity of combined system(v) = ? it is a combined system because the girl came on the cart so their velocity would be same. m1u1 +m2u2 = m1v +m2v 40*5 + 10*1 = (m1+m2)v 200 + 10 = (40+10)v 210 = 50v v = 21/5m/s = 4.2 m/s
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Answer:
4.2 m/s
Explanation:
Explanation: law of conservation of momentum. mass of girl(m1) = 40 kg initial velocity of girl(u1) = 5m/s mass of cart(m2) = 10 kg initial vel. of cart(u2) = 1m/s final velocity of combined system(v) = ? it is a combined system because the girl came on the cart so their velocity would be same. m1u1 +m2u2 = m1v +m2v 40*5 + 10*1 = (m1+m2)v 200 + 10 = (40+10)v 210 = 50v v = 21/5m/s = 4.2 m/s