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.
You might be interested in
Explanation:
Given that,
Charge acting on the object,
Force acting on the object, (in downward direction)
(a) The electric force acting in the electric field is given by :
E is the electric field
E = 4.75 N/C
The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.
(b) The charge on the proton is,
The force acting on the proton is :
If the charge on the proton is positive, the force on the proton is in upward direction.
Hence, this is the required solution.