Answer:
1) v = -0.89 m/s
2) v = 36.5 m/s
Explanation:
1) Given
The mass of the boy, M = 45 Kg
The mass of the skateboard, m = 2 Kg
The initial velocity of the boy and skateboard is zero
According to the law of conservation of linear momentum
MV + mv = 0
V = - mv/M
Substituting the given values in the above equation
V = -2 x 20/ 45
= -0.89 m/s
Hence, the velocity of the boy, v = -0.89 m/s
2) Given
The mass of the boy, M = 47 Kg
The mass of the skateboard, m = 8 Kg
The initial combined velocity of the boy and skateboard, u = 4.2 m/s
The boy jumped off from the skateboard with velocity, V = -1.3 m/s
According to the law of conservation of momentum,
MV + mv = (M + m)u
v = [(M + m)u - MV] / m
Substituting the given values,
v = [(47 + 8)4.2 - 47(-1.3)] / 8
= 36.5 m/s
Hence, the velocity of the skateboard, v = 36.5 m/s
