Answer:
2 m/s
Explanation:
Parameters given:
Mass of first skateboard, m = 3 kg
Initial speed of first skateboard, u = 4 m/s
Mass of second skateboard, M = 1 kg
Initial speed of second skateboard, U = 0 m/s
Final speed of second skateboard, V = 6 m/s
Using the principle of the conservaton of momentum, the total initial momentum is equal to the total final momentum.
Momentum is the product of mass and velocity. This implies that:
m*u + M*U = m*v + M*V
(3*4) + (1*0) = (3*v) + (1*6)
12 + 0 = 3v + 6
=> 3v = 12 - 6
3v = 6
v = 6/3 = 2 m/s
The final speed of the 3 kg skateboard is 2 m/s
Answer:
2m/s
Explanation:
Using the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of their momentum after collision.
Momentum = mass × velocity.
Before collision:
Momentum of 3kg body moving at 4m/s = 3×4
= 12kgm/s
Momentum of 1kg skateboard initially at rest = 1×0
= 0kgm/s (the velocity 0m/s because the body is initially at rest)
After collision:
1-kg skateboard moves off at 6m/s. Its momentum after collision will be;
1×6 = 6kgm/s
For the 3kg body moving with an unknown velocity v, its final momentum will be;
3×v = 3v
Applying the law as stated above;
12+0 = 6+3v
12 =6+3v
12-6 = 3v
6 = 3v
v = 6/3
v = 2m/s
This means that the new velocity of the 3kg skateboard after collision is 2m/s
