before colliding the momentum of block A is -100 kg*m/s, and block B is -150 kg*m/s. after, block A has a momentum -200 kg*m/s.
1 answer:
The momentum of block B after the collision is -50 kg m/s.
Explanation:
We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:
where:
is the momentum of block A before the collision
is the momentum of block B before the collision
is the momentum of block A after the collision
is the momentum of block B after the collision
Solving for , we find:
So, the momentum of block B after the collision is -50 kg m/s.
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Answer:
b. 0.20 m/s.
Explanation:
Given;
initial mass, m = 0.2 kg
maximum speed, v = 0.3 m/s
The total energy of the spring at the given maximum speed is calculated as;
K.E = ¹/₂mv²
K.E = 0.5 x 0.2 x 0.3²
K.E = 0.009 J
If the mass is changed to 0.4 kg
¹/₂mv² = K.E
mv² = 2K.E
Therefore, the maximum speed is 0.20 m/s