Perfectly elastic collisions in one dimension – problems and solutions

1. A 200-gram ball, A, moving at a speed of 10 m/s strikes a 200-gram ball, B, at rest. What is the speed of ball A and ball B after the collision?

Known :

Mass of ball A (m_A) = 200 gram = 0.2 kg

Mass of ball B (m_B) = 200 gram = 0.2 kg

Speed of ball A before collision (v_A) = 10 m/s

Speed of ball B before collision (v_B) = 0

Wanted: speed of ball A (v_A’) and speed of ball B (v_B’) after a collision

Solution :

v_A’ = v_B = 0

v_B’ = v_A = 10 m/s

2. Two equal-mass objects approach each other, collide and then bounce off. The speed of mass A before the collision is 8 m/s and speed of mass B before the collision is 12 m/s. What is the magnitude and direction of the velocity of mass A and mass B after the collision!

Known :

Mass A (m_A) = m

Mass B (m_B) = m

Speed of mass A before collision (v_A) = -8 m/s

Speed of mass B before collision (v_B) = 12 m/s

The plus and minus sign indicates that the objects move in opposite direction.

Wanted: the magnitude and direction of the velocity of mass A (v_A’) and mass B (v_B’) after a collision

Solution :

In a perfectly elastic collision, if the objects of equal mass then the speed of A after the collision (v_A’) = the speed of B before collision (v_B) and the speed of B after the collision (v_B’) = the speed of A before collision (v_A).

If before collision, A moving rightward and B moving leftward, then, after collision, A moving leftward and B moving rightward.

The objects have equal mass and the collision is perfectly elastic so the both objects exchanged speed.

v_A’ = -v_B = -12 m/s

v_B’ = v_A = 8 m/s

The plus and minus sign indicates that the objects move in opposite direction.

3. A 2-kg object, A, moving at a speed of 10 m/s strikes a 4-kg ball, B, moving at a speed of 20 m/s. If the collision is perfectly elastic, what is the speed of object A and object B after the collision?

Known :

Mass A (m_A) = 2 kg

Mass B (m_B) = 4 kg

Speed of object A before collision (v_A) = 10 m/s

Speed of object B before collision (v_B) = 20 m/s

Wanted : the speed of object A (v_A’) and the speed of object B (v_B’) after collision

Solution :

In perfectly elastic collision, if the objects have equal mass and approach each other, the speed of the object after collision calculated using this formula :

Speed of object A after collision :

4. A 100-gram moving at 20 m/s strikes a wall perfectly elastic collision. What is the magnitude and direction of object’s velocity after collision.

Known :

Mass (m_A) = 100 gram = 0.1 kg

Mass of wall (m_B) = infinity

The speed of mass before collision (v_A) = 20 m/s

The speed of mass after collision (v_B) = 0

Wanted : the magnitude and direction of velocity after collision

Solution :

If v_B = 0, m_A very small and m₂ very big then v_A’ = -v_A and v₂’ = 0.

The plus and minus sign indicates that the objects move in the opposite direction.

5. Two balls, A with a mass of 2-kg and ball B with a mass of 5-kg, before and after the collision are shown in the figure below. If the collision is perfectly elastic, what is the speed of ball A before the collision?

Known :

Mass of ball A (m_A) = 2 kg

Mass of ball B (m_B) = 5 kg

Speed of ball B before collision (v_B) = 2 m/s

Speed of ball A after collision (v_A‘) = 5 m/s

Speed of ball B after collision (v_B‘) = 4 m/s

Both balls are move to the right so their speed signed positive.

Wanted : Speed of ball A before collision (v_A)

Solution :

m_A v_A + m_B v_B = m_A v_A‘ + m_B v_B‘

2(v_A) + (5)(2) = (2)(5) + (5)(4)

2(v_A) + 10 = 10 + 20

2(v_A) + 10 = 30

2(v_A) = 30 – 10

2(v_A) = 20

v_A = 20/2

v_A = 10 m/s

6. Object A and B travel along horizontal plane, as shown in figure below. If the collision is perfectly elastic and the speed of object B after collision is 15 m/s¹, what is the speed of object A after collision.

Known :

Mass of object A (m_A) = 5 kg

Mass of object B (m_B) = 2 kg

Speed of object A before collision (v_A) = 12 m/s

Speed of object B before collision (v_B) = 10 m/s

Speed of object B after collision (v_B‘) = 15 m/s

Both objects, before and after collision, move to right so their velocity signed positive.

Wanted : Speed of object A after collision (v_A‘)

Solution :

m_A v_A + m_B v_B = m_A v_A‘ + m_B v_B‘

(5)(12) + (2)(10) = (5)v_A‘ + (2)(15)

60 + 20 = (5)v_A‘ + 30

80 = (5)v_A‘ + 30

80 – 30 = (5)v_A‘

50 = (5)v_A‘

v_A‘ = 50/5

v_A‘ = 10 m/s

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