A solid ball rolls along a frictionless plane. It is 4 kilograms and is moving at a velocity of 15 meters per second. If it has a completely elastic collision and transfers all of its momentum to a stationary solid ball of 14 kilograms, what will be the velocity of the second ball?
The momentum of the first ball is 4*15 kg m/s
This is equal to the momentum of the 14 kg ball after it starts moving:
4*15 = 14*v
where v is the velocity of the second ball
I suspect your teacher did not realize what he wrote: commpletely elastic, and transfers all of its momentum....
Rotational energy, and rotational momentum being conserved is an issue here, I think. Nothing on the size of either ball is given...so my guess is your teacher MEANT to say, the collision in translational motion was elastic, and all translational momentum was transfered. Now in reality, it would be a feat to do this with balls of different mass, as any pool player can attest, but they don't play on a frictionaless table, where balls theoritically move without rolling.
So considering translational motion only.
KE before= KE after
1/2 4 * 15^2=1/2 14 vf^2 give you the velocity of the second ball.
Now if you consider momentum
momentum before=momentum after
4*15=14*Vf
Notice it is a different answer than in the KE solution.
Thus, as pool players recognize, it is not possible. Maybe your teacher had something in mind here, but it escapes me.
To find the velocity of the second ball after the elastic collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v), expressed as p = m * v.
Let's denote the initial velocity of the first ball as v1, the mass of the first ball as m1, and the velocity of the second ball after the collision as v2.
Given:
Mass of the first ball (m1) = 4 kg
Velocity of the first ball (v1) = 15 m/s
Mass of the second ball (m2) = 14 kg
Velocity of the second ball (v2) = ?
Using the principle of conservation of momentum, we can write the equation:
Initial momentum = Final momentum
(mass of first ball * initial velocity of first ball) = (mass of first ball * final velocity of first ball) + (mass of second ball * final velocity of second ball)
(4 kg * 15 m/s) = (4 kg * v1) + (14 kg * v2)
Now we can solve for v2:
60 kg·m/s = 4 kg * 15 m/s + 14 kg * v2
60 kg·m/s = 60 kg·m/s + 14 kg * v2
60 kg·m/s - 60 kg·m/s = 14 kg * v2
0 kg·m/s = 14 kg * v2
Therefore, the final velocity of the second ball (v2) after the elastic collision is 0 m/s. It comes to rest since all the momentum of the first ball is transferred to it.