A ball with mass m=1kg and speed vo=5m/sec elastically collides with a stationary, identical ball (all resting on a frictionless surface so gravity is irrelevant). A student measures the top ball emerging from the collision at at speed vt=4m/sec at an angle ϴt ≈ 37°.
a) Find the speed vb of the other ball.
b) Find the angle ϴb of the other ball.
c) What does ϴt+ϴb add up to?
To find the answers to these questions, we can apply the conservation of momentum and conservation of kinetic energy principles.
a) To find the speed vb of the other ball, we use the conservation of momentum principle. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision:
The momentum of the first ball (with mass m) can be calculated as momentum1 = mass * velocity = m * vo.
Since the second ball is stationary, its momentum is zero.
After the collision:
The momentum of the first ball can be calculated as momentum1' = mass * velocity = m * vt.
The momentum of the second ball is given by momentum2' = mass * velocity = m * vb.
Using the conservation of momentum principle, we have:
momentum1 + momentum2 = momentum1' + momentum2'
m * vo + 0 = m * vt + m * vb
Simplifying the equation:
vo = vt + vb
5 m/s = 4 m/s + vb
Solving for vb:
vb = vo - vt
vb = 5 m/s - 4 m/s
vb = 1 m/s
Therefore, the speed of the other ball (vb) is 1 m/s.
b) To find the angle ϴb of the other ball, we need to apply the conservation of kinetic energy principle. According to this principle, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Before the collision:
The kinetic energy of the first ball is given by KE1 = (1/2) * mass * (velocity)^2 = (1/2) * m * (vo)^2.
The kinetic energy of the second ball is zero since it is stationary.
After the collision:
The kinetic energy of the first ball is given by KE1' = (1/2) * m * (velocity)^2 = (1/2) * m * (vt)^2.
The kinetic energy of the second ball is given by KE2' = (1/2) * m * (velocity)^2 = (1/2) * m * (vb)^2.
Using the conservation of kinetic energy principle, we have:
KE1 + KE2 = KE1' + KE2'
(1/2) * m * (vo)^2 + 0 = (1/2) * m * (vt)^2 + (1/2) * m * (vb)^2
Simplifying the equation:
(vo)^2 = (vt)^2 + (vb)^2
(5 m/s)^2 = (4 m/s)^2 + (vb)^2
25 m^2/s^2 = 16 m^2/s^2 + (vb)^2
Solving for (vb)^2:
(vb)^2 = 25 m^2/s^2 - 16 m^2/s^2
(vb)^2 = 9 m^2/s^2
Taking the square root of both sides:
vb = √(9 m^2/s^2)
vb = 3 m/s
Therefore, the speed of the other ball (vb) is 3 m/s.
c) To find ϴt+ϴb, we simply add the angles together:
ϴt+ϴb ≈ 37° + ϴb
Since we have already determined that ϴb is unknown, we cannot find the exact sum.
In conclusion:
a) The speed of the other ball (vb) is 1 m/s.
b) The angle of the other ball (ϴb) cannot be determined without additional information.
c) The sum of ϴt+ϴb cannot be determined without the value of ϴb.
a,b. M1*V1 + M2*V2 = M1*V3 + M2*V4.
1*5 + 1*0 = 1*4[37o] + 1*V4.
5 = 4[37o] + V4, V4 = 5-4[37O],
V4 = 5-4*Cos37-4*sin37 = 5-3.19-2.41i = 1.81 - 2.41i = 3.01m/s[-53.1o] = 3.01m/s[53.1o] S. of E. = Velocity and Direction.
c. A1+A2 = 37 + 53.1 = 90.1o.