Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts
1 answer:
Answer:
0.304 m/s2
Explanation:
If the first child is pushing with a force of 69N to the right and the 2nd child is pushing with a force of 91N to the left. Then the net pushing force is 91 - 69 = 22 N to the left. Subtracted by 15N friction force then the system of interest is subjected to F = 7 N net force tot he left.
We can use Newton's 2nd law to calculate the net acceleration of the system
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Answer:
Explanation:
Since the surface is frictionless, momentum will be conserved. If the bullet of mass has an initial velocity
and a final velocity
and the block of mass
has an initial velocity
and a final velocity
then the initial and final momentum of the system will be:
Since momentum is conserved, , which means:
We know that the block is brought to rest by the collision, which means and leaves us with:
which is the same as:
Considering the direction the bullet moves initially as the positive one, and writing in S.I., this gives us:
So kinetic energy of the bullet as it emerges from the block will be: