A 53.4 kg girl is at rest on 3.55 kg skateboard. She jumps forward at 1.32 m/s. What is the velocity of the skateboard afterword
?
1 answer:
M = mass of the girl = 53.4 kg
m = mass of skateboard = 3.55 kg
V' = velocity of the combination of girl and skateboard before she jumps forward = 0
V = velocity of the girl forward = 1.32
v = velocity of the skateboard afterward = ?
Using conservation of momentum
(M + m) V' = MV + m v
inserting the values
(53.4 + 3.55) (0) = (53.4) (1.32) + (3.55) v
v = - 19.86
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Answer: -3.49 m/s (to the south)
Explanation:
This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum must be equal to the final momentum
, and taking into account this is aninelastic collision:
Before the collision:
(1)
After the collision:
(2)
Where:
is the mass of the car
is the velocity of the car, directed to the north
is the mass of the truck
is the velocity of the truck, directed to the south
is the final velocity of both the car and the truck
(3)
(4)
Isolating :
(5)
(6)
Finally:
The negative sign indicates the direction of the velocity is to the south