Any help with parts a-e are appreciated. A motionless object, mi of mass 1.5 kg

Question

Any help with parts a-e are appreciated.

A motionless object, mi of mass 1.5 kg is at a height h_1i of 2.9 m above motionless object m_2 of mass 3.0 kg. Both objects can swing freely from massless ropes. You might choose to use the following equation for elastic collisions: v_1, I - v_2, I = -(v_1, f - v_2, f) If object m_1 is released so it swings down and hits object m_2 with a perfectly horizontal velocity, what is the speed of object m_1, immediately before it has an elastic collision with object m_2? What is the velocity of m_1 just after the elastic collision? What is the velocity of m_2 just after the elastic collision? What is the height to which m_1 rises after the elastic collision? What is the height to which m_2 rises after the elastic collision?

Explanation / Answer

m1 = 1.5 kg , h =2.9 m , m2 = 3kg

From conservation of energy of m1 objcet

kinetic energy = potential energy

mgh = 1/2(mu^2)

u =[2gh]^1/2 = [2*9.8*2.9]^1/2

velcoity of first object before collision u = 7.54 m/s

(b) From conservation of momentum

m1u1+m2u2 =m1v1+m2v2

1.5*7.54 +0 = 1.5v1+3v2

11.31 = 1.5 v1 +3v2 ... (1)

In elastic collision

v2 -v1 = u1 -u2

v2 - v1 = 7.54 ..(2)

from (1) and (2)

(b) v1 = -2.51 m/s

(c) v2 = 5.03 m/s

(d) v1 = 2.51 m/s

h = v1^2/2g = (2.51*2.51)/(2*9.8)

h = 0.32 m

(e) v2 = 5.03 m/s

h = v2^2/2g = (5.03*5.03)/(2*9.8)

h = 1.291 m

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