Basic Physics

# Impulse, momentum, collisions – problems and solutions

1. A 40-gram rubber bullet shot horizontally to the wall, as shown in the figure below. The bullet is reflected at the same speed. What is the change in momentum of the ball?

Known : Mass (m) = 40 gram = 0.04 kg

Initial velocity (vo) = – 50 m/s

Final velocity (vt) = 50 m/s

Direction of ball’s displacement (direction of velocity) is opposite so that the initial velocity and the final velocity have opposite signs.

Wanted : The change in momentum of ball

Solution :

The change in momentum :

Δp = m (vt – vo) = (0.04)(50 – (-50)) = (0.04)(50 + 50)

Δp = (0.04)(100) = 4 N.s

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2. A 75-gram rubber ball thrown horizontally to hit the wall, as shown in figure below. The ball is reflected at the same speed. What is the impulse.

Known :

Mass of ball (m) = 75 gram = 0.075 kg Initial velocity (vo) = -20 m/s

Final velocity (vt) = 20 m/s

Direction of ball’s displacement (direction of velocity) is opposite so that the initial velocity and the final velocity have opposite signs.

Wanted : Impulse

Solution :

Impulse = The change in momentum

I = Δp

I = m (vt – vo) = 0.075 (20 – (-20)) = 0.075 (20 + 20) = 0.075 (40)

I = 3 N s = 3 Newton second

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Perfectly elastic collision

3. Two balls, A and B, approach each other along a horizontal plane. Velocity of ball A, vA = 4 m/s and velocity of ball B, vB = 6 m/s. The collision is perfectly elastic. The velocity of object B after collision is 4 m/s. What is the velocity of ball A after collision.

Known :

Mass of ball A (mA) = m Mass of ball B (mB) = m

Velocity of ball A before collision (vA) = 4 m/s

Velocity of ball B before collision (vB) = 6 m/s

Velocity of ball B after collision (vB’) = 4 m/s

Wanted: Velocity of ball A after a collision (vA’)

Solution :

Mass of balls are same, so that :

Velocity of ball A before collision (vA) = velocity of ball B after collision (vB’) = 4 m/s

Velocity of ball B before collision (vB) = velocity of ball A after collision (vA’) = 6 m/s

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Partially Elastic Collisions

4. Two objects with the same mass move in a straight line approaching each other, as shown in the figure below. If v2 is the velocity of the object 2, after the collision to the right at 5 m.s-1, then the velocity of the object (1) after the collision is…

Known :

Object’s mass = m Velocity of object 1 before collision (v1) = 8 m/s

Velocity of object 2 before collision (v2) = 10 m/s

Velocity of object 2 after collision (v2‘) = 5 m/s

Wanted : Velocity of object 1 after collision (v1‘)

Solution :

m1 v1+ m2 v2 = m1 v1’ + m2 v2

m (v1 + v2) = m (v1’ + v2’)

v1 + v2 = v1’ + v2

8 + 10 = v1’ + 5

18 = v1’ + 5

v1’ = 18-5

v1’ = 13 m/s

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5. Object A with mass of 4 kg and object B with mass of 5 kg approach each other along a horizontal plane as shown in figure below. After collision, velocity of ball A = 4 m/s and velocity of ball B = 2 m/s. What is the velocity of ball B before collision.

Known :

Mass of object A (mA) = 4 kg Mass of object B B (mB) = 5 kg

Velocity of object A before collision (vA) = 6 m/s

Velocity of object A after collision (vA’) = 4 m/s

Velocity of object B after collision (vB’) = -2 m/s

Wanted : Velocity of ball B before collision (vB)

Solution :

mA vA + mB vB = mA vA’ + mB vB

(4)(6) + (5)(-2) = (4)(4) + (5)(vB’)

24 – 10 = 16 + 5(vB’)

14 – 16 = 5 (vB’)

-2 = 5 (vB’)

vB’ = -2/5

vB’ = -0.4

Minus sign indicates that direction of ball after collision is opposite with the direction of ball before collision.

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Inelastic Collisions

6. A 2-kg ball and a 1-kg ball approach each other along a horizontal plane, as shown in figure below.

Velocity of ball 1, v1 = 2 ms-1 and velocity of ball 2, v2 = 4 ms-1. After the collision, both stick together. What is the velocity after the collision?

Known :

Mass of ball 1 (m1) = 2 kg Mass of ball 2 (m2) = 1 kg

Velocity of ball 1 before collision (v1) = 2 m/s

Velocity of ball 2 before collision (v2) = -4 m/s

Direction of ball’s displacement (direction of velocity) is opposite so that velocity of ball 1 and velocity of ball 2 have opposite signs.

Wanted : Velocity after collision (v’)

Solution :

m1 v1 + m2 v2 = (m1 + m2) v’

(2)(2) + (1)(-4) = (2 + 1) v’

4 – 4 = (3) v’

0 = (3) v’

v’ = 0

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