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A 100 g ball moving to the right at 4.0 m/s collides head-on with a 200 g ball that is moving to the left at 3.0 m/s.

a. If the collision is perfectly elastic, what are the speed and direction of each ball after the collision?

b. If the collision is perfectly inelastic, what are the speed and direction of the combined balls after the collision?

Short Answer

Expert verified

a) Speed of First ball v1f = -5.33 m/s (move towards Left)

Speed of second ball v2f = 1.67 m/s (move towards Right)

b) Speed of combined balls vf = -0.67 m/s (move towards Left)

Step by step solution

01

Step 1. Given Information is :Mass of first object m1 = 100g = 0.1 kgVelocity of first object v1 = 4 m/sMass of second object m2 = 200g = 0.2 kgVelocity of second object v2 = -3 m/s (Negative as it is moving towards Left)

We need to find out :

a) The speed and direction of each ball after the collision in case of Elastic collision

b) The speed and direction of the combined balls after the collision in case of Inelastic Collision

02

Step 2. Part a) For elastic collision, Using Galilean transformation of velocities,

Velocity of object 1 in lab = v1L = 4 m/s

Velocity of object 2 in lab = v2L= -3 m/s

If velocity of object 2 in frame = v2M = 0

So velocity of frame w.r.t lab = vML = -3 m/s

Velocity of lab w.r.t frame = vLM = 3 m/s

Therefore,

Velocity of object 1 in frame M = v1M = v1L + vLM = 4 + 3

= 7 m/s

Using formula in case of elastic collision for velocities,

v1Mf=m1-m2m1+m2v1M=0.1-0.20.1+0.27=-2.33m/sv2Mf=2m1m1+m2v1M=20.10.1+0.27=4.67m/s

Converting them into Lab frame,

v1Lf = v1Mf + vML

= -2.33 + (-3)

= -5.33 m/s

v2Lf = v2Mf + vML

= 4.67 - 3

= 1.67 m/s

03

Step 3. Part b) The speed and direction of the combined balls after the collision in case of inelastic collision

We know that in case of Inelastic collision,

(m1+m2)vf=m1v1+m2v2(0.1+0.2)vf=0.14+0.2(-3)0.3vf=0.4-0.6=-0.2vf=-0.20.3=-0.67m/s

(Negative sign indicates that they will move together towards Left)

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