A 100 g ball moving to the right at 4.0 m/s collides head-on with a 200g ball th

Question

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

Part A & B

If the collision is perfectly elastic, what are the speeds of each ball after the collision?

Express your answer with the appropriate units.

(vfx)1 =

(vfx)2 =

Part C

What is the direction of 100-g ball after the collision?

Part D

What is the direction of 200-g ball after the collision?

Part E

If the collision is perfectly inelastic, what is the speed of the combined balls after the collision?

Express your answer with the appropriate units

Part F

What is the direction of the combined balls after the collision?

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

Part A & B

If the collision is perfectly elastic, what are the speeds of each ball after the collision?

Express your answer with the appropriate units.

(vfx)1 =

(vfx)2 =

Part C

What is the direction of 100-g ball after the collision?

upward downward to the right to the left

Part D

What is the direction of 200-g ball after the collision?

upward downward to the right to the left

Part E

If the collision is perfectly inelastic, what is the speed of the combined balls after the collision?

Express your answer with the appropriate units

vfx =

Part F

What is the direction of the combined balls after the collision?

upward downward to the right to the left

Explanation / Answer

Given: m1= 100 g , v1= 4 m/s

m2= 200 g, v2= 3 m/s

A) if collision is perfectly elastic, then momentum and kinetic energy will be conserved.

Choosing right as positive

Initial Momentum: m1v1-m2v2 [ second ball is moving to left Ans we chooe right as positive direction]

Initial momentum : (0.100×4)-(0.200×3)= -0.2 kg m/s

Final momentum: (0.100× Vf1 + 0.200× Vf2

By equating:

-0.2 = 0.100 Vf1+ 0.200 Vf2

Vf2= -1.5 Vf1

Now Initial kinetic energy = Final kinetic energy

(1/2) (0.100)(42)+(1/2)(0.200)(32)= (1/2)(0.100)(Vf12)+(1/2)(0.200)(Vf22)

By putting value of Vf2

3.4 = 0.100(Vf12)+(0.200)(1.5Vf1)2

Solving

Vf1= 2.49 m/s and Vf2= - 1.5 Vf1= -3.73 m/s

B) direction of 100 g ball is to right

D)Direction of 200 g ball is to the left

E) if collision is perfectly inelastic , then

Momentum is conserved and masses will stick to each other

m1v1-m2v2= (m1+m2)vf

(0.100)(4)-(0.200)(3) = (0.100+0.200)Vf

Vf= -0.67 m/s

Negative sign indicates that , direction will be to the left

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.