Newton\'s Cradle (demonstrated in lecture) consists of 5 steel balls, individual

Question

Newton's Cradle (demonstrated in lecture) consists of 5 steel balls, individually held by vertical strings. Collisions that occur between the balls are essentially elastic and hence demonstrate two conservation laws. If the end ball is moved to a position as shown in the diagram, what is its change in gravitational potential energy from when it was hanging vertically? The ball in A) is released from rest at angle theta. After it makes a collision only the right most ball moves. Determine its velocity. For parts C), D), and E) assume that the 4 balls hanging vertically are taped together. Determine the velocity of this group of 4 balls after a collision with the ball that is let go form rest all the angle theta. Determine the velocity of the ball (that was released) after the collision. Determine the total momentum of the system of 5 balls after the collision.

Explanation / Answer

A) Gravitational potential energy is PE = m*g*h = m*g*l(1-cos(55)) = (75*10^-3*9.8*0.25*(1-cos(55))) = 0.0783 J

B) Using law of conservation of energy

m*g*l(1-cos(55)) = 0.5*m*v^2

m cancels

9.8*0.25*(1-cos(55)) = 0.5*v^2

v = 1.45 m/sec

C) Using law of conservation of energy

m*g*l(1-cos(55)) = 0.5*4*m*v^2

m cancels

9.8*0.25*(1-cos(55)) = 0.5*4*v^2

v = 0.73 m/sec

D) Using law Conservation of momentum

m*Vo = (m*V1) + (4*m*V2)

Vo = sqrt(2*g*l*(1-cos(55))) = sqrt(2*9.8*0.25*(1-cos(55)))

Vo = 1.44 m/sec

then

m*Vo = (m*V1) + (4*m*V2)

m cancels

1.44 = V1+(4*0.73)
V1 = 1.48 m/sec opposite to the original direction

E) total momentum is m*vo = 75*10^-3*1.44 = 0.108 kg m/sec

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