Question

Identical twins, each with mass 55 kg, are on ice skates and at rest on a frozen lake, which may be taken as frictionless. Twin A is carrying a backpack and she throws it horizontally at 3.8 m/s to Twin B. After throwing the backpack, twin A gains a velocity of 1.4.

Neglecting any gravity effects, what are the subsequent speeds of Twin B in m/s.

Answer 1

Final Answer:-

Velocity of twin B is 1.03 m/s.

Solution:- when twin A throws the bag then,

according to conservation of momentum,

initial momentum = final momentum

mA*viA + M * Vi = mA * vfA + M * Vf

as initially both are on rest so,

0 + 0 = 55 * (-1.4) + M * 3.8

as per the Newton's third law, their velocities will be in opposite directions.

M = 55*1.4/3.8

M = 20.26 kg

Now will calculate the velocity of B when she receive the bag.

similarly applying conservation of momentum,

mB * viB + M * Vi = mB * vfB + M * Vf

now before recieving the bag the velocity of B is zero and when he will catch the bag, the final velocity of the bag will be same as velocity of B,

0 + (20.26) * 3.8 = (55 + 20.26) * vfB

solving for vfB,

vfB = 76.98/75.26

vfB = 1.03 m/s

Thus the velocity of the B is 1.03 m/s.