As shown in the figure below, object m1 = 1.75 kg starts at an initial height h1

Question

As shown in the figure below, object m1 = 1.75 kg starts at an initial height h1i = 0.260 m and speed v1i = 4.00 m/s, swings downward and strikes (in an elastic collision) object m2 = 4.80 kg which is initially at rest. Determine the following.

(a) speed of m1 just before the collision. _______ m/s

(b) velocity (magnitude and direction) of each ball just after the collision (Assume the positive direction is toward the right. Indicate the direction with the sign of your answer.)

_______ m/s (m1)

_______ m/a (m2)

(c) height to which each ball swings after the collision (ignoring air resistance)

_______ m (m1)

_______ m (m2)

Explanation / Answer

Given that

An object of mass m1 = 1.75 kg

The object of mass m1 starts at an initial height h1i = 0.260 m

The speed of the first object and speed v1i = 4.00 m/s,

swings downward and strikes (in an elastic collision) object m2 = 4.80 kg which is initially at rest(v2i) =0

The initial velocity if the ball is given by

Eo =(1/2)m1v1i2+m1gh =0.5*1.75kg*(4m/s)2+(1.75kg)(9.81m/s2)(0.260m) =14+4.4635m=18.4635J

Now final energy of the ball just before the collision is given by

Ef =(1/2)m1vf2 =18.4635J

vf =Sqrt(2*18.4635/1.75) =4.593m/s

b)

When it is perfectly elastic collision and the second body is at rest v2i =0 then

v1f =(m1-m2)vi1/(m1+m2) =(1.75kg-4.80kg)(4m/s)/(1.75+4.80) =-12.2/6.55=-1.8625m/s (negative sign indicates the ball one is moving in negative x-direction)

v2f =2m1v1i/(m1+m2) =2*1.75kg*4m/s/6.55 =14/6.55 =2.1374m/s (positive x -direction)

c)

Now from the conservation of energy

m1gh1 =(1/2)m1v1f2 ====>h1 =(1/2)m1v1i2 /m1g =0.5*(-1.8625)2/9.81 =0.1768m

Similarly

m2gh2 =(1/2)m2v2f2 =====>h2 =(1/2)m2v2f2/m2g =0.5(2.1374)2/9.81 =0.2328m

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