physics help! An elastic ball (A) with mass M is released from a horizontal posi

Question

physics help!

An elastic ball (A) with mass M is released from a horizontal position, connected via a massless string (length R)to a rod. When the ball reaches the bottom, it collides with another elastic ball (B) of mass m, aslo connected via a massless string (length R) to the rod. a)What is the velocity of ball B (m) after the collision? You answer should be expressed in terms of m, M and g and R(5pt) b)What is magnitude of the tension force (T) in the string connected to the ball B at the moment right after the collision? You answered should be expressed in terms of M, m and g(5pt) c) In order for the ball B to reach all the way to the top without falling, what is the minimum value of Your answer should be a pure number. (10 pt) M M m

Explanation / Answer

PE1 of ball A = MgR

KE of ball A before collision = 1/2*M*v1_a^2

1/2*M*v1_a^2 = MgR

v1_a = sqrt (2gR)

Momentum conservation before and after collision:

M*v1_a + 0 = M*v2_a + m*v2_b

M*sqrt (2gR) = M*v2_a + m*v2_b..............1

KE conservation before and after collision:

1/2*M*v1_a^2 + 0 = 1/2*M*v2_a^2 + 1/2*m*v2_b^2

2MgR = M*v2_a^2 + m*v2_b^2...................2

Solving 1 and 2,

2MgR = M*(sqrt (2gR) - (m/M)*v2_b)^2 + m*v2_b^2

2MgR = M*[2gR + (m/M)^2 * v2_b ^2 - 2*sqrt(2gR)*(m/M) *v2_b] + m*v2_b^2

2*sqrt(2gR)*(m/M) *v2_b = (m/M)^2 * v2_b ^2 + (m/M)*v2_b^2

2*sqrt(2gR) = [1 + (m/M)]*v2_b

v2_b = 2*sqrt(2gR) / [1 + (m/M)]

b)

Centrifugal force = m*v2_b^2 / R

= 8mg / [1 + (m/M)]^2

T = mg + 8mg / [1 + (m/M)]^2

c)

KE of b = 1/2*m*v2_b^2

PE of b at top = mg*(2R)

Equating both, mg*2R = 1/2 *m*v2_b^2

4gR = 4*(2gR) / [1 + (m/M)]^2

[1 + (m/M)]^2 = 2

m/M = 0.414

M/m = 2.414

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