A 98-kg fullback is running along at 8.6 m / s when a 76-kg defensive back running in the same direction at 9.8 m / s jumps on h
1 answer:
The total momentum before and after the collision must be conserved.
The total momentum before the collision is:
where m1 and m2 are the masses of the two players, and
and 
their initial velocities. Both are considered with positive sign, because the two players are running toward the same direction.
The final momentum is instead
because now the two players are moving together with a total mass of (m1+m2) and final speed vf.
By requiring that the momentum is conserved
we can calculate vf, the post-collision speed:
and the direction is the same as the direction of the players before the collision.
The total momentum before the collision is:
where m1 and m2 are the masses of the two players, and
The final momentum is instead
because now the two players are moving together with a total mass of (m1+m2) and final speed vf.
By requiring that the momentum is conserved
we can calculate vf, the post-collision speed:
and the direction is the same as the direction of the players before the collision.
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Answer:
The kinetic energy of the particle will be 12U₀
Explanation:
Given that,
A particle is launched from point B with an initial velocity and reaches point A having gained U₀ joules of kinetic energy.
Constant force = 12F
According to question,
The kinetic energy is
....(I)
Constant force = 12F
A resistive force field is now set up ,
Resistive force is given by,
When the particle moves from point B to point A then,
We need to calculate the kinetic energy
Using formula for kinetic energy
Put the value of
Now, from equation (I)
Hence, The kinetic energy of the particle will be 12U₀.
