Question

1- Block 1 (mass 2.0 kg) is moving to right at speed of 5.0 m/s. It collides with block 2 (mass 3.0 kg) that is moving to the right at speed of 3.0 m/s.

a) What is the speed of the blocks if the collision is completely inelastic? How much kinetic energy is lost? (3.8 m/s, 2.4 J)

b) What is the final speed of the blocks if the collision is elastic? (v1 = 2.6 m/s, v2 = 4.6 m/s)

2 - A solid uniform sphere of mass 5.0 kg and diameter 40 cm, rolls down a hill without slipping. If the velocity of the sphere downhill is 2.0 m/s.

a) what is the height of the hill? (29 cm)

b) What is the angular momentum of the sphere downhill? (0.8 kg*m/s)

I know the answers I just need to know how to get to those answers!

Answer 1

a.)

The objects stick together in a perfectly inelastic collision, and so their final will be same.

now by momentum conservation,

Pi = Pf

m1*v1 + m2*v2 = (m1+m2)*V

here, m1 = 2.0 kg

m2 = 3.0 kg

v1 = +5.0 m/s

v2 = +3.0 m/s

V = final speed =??

So, 2*5 + 3*3 = (2+3)*V

V = 19/5

V = 3.8 m/s

now loss in kinetic energy is given by,

dKE = KEf - KEi

dKE = (0.5*m1*v1^2 + 0.5*m2*v2^2) - (0.5*m1*V^2 + 0.5*m2*V2)

dKE = (0.5*2*5^2 + 0.5*3*3^2) - (0.5*(2+3)*3.8^2)

dKE = 2.4 J

b.)

In a perfectly elastic collision
Using momentum conservation
Pi = Pf
m1V1i + m2V2i = m1V1f + m2*V2f
given that m1 = 2.0 kg

m2 = 3.0 kg

V1i = +5.0 m/s

V2i = +3.0 m/s

So, 2*5 + 3*3 = 2*V1f + 3*V2f
2*V1f + 3*V2f = 19
Now In elastic collisions,
V1f - V2f = V2i - V1i
V1f - V2f = +3 - (+5)
V1f - V2f = -2

by multiply with 2,

2*V1f - 2*V2f = -4
Now Solving both bold equation
subtract both of them,

3*V2f + 2*V2f = 19 + 4

V2f = 23/5 = 4.6 m/s

So, V1f = -2 + V2f = -2 + 4.6

V1f = 2.6 m/s

Since rest of the questions are not related, So please ask them as a new question, I will be happy to help.

Please upvote.