Question

a body of mass 6kg travelling with a velocity 10m/s collides head on and elastically with a body of mass 4kg travelling at a speed of 5m/s in opposite direction. the velocity of the second body after collision is

Answer 1

m₁ = 6kg
u₁ = 10m/s
m₂ = 4 kg
u₂ = -5 m/s
v₂ = ?

$v_2= \frac{u_2(m_2-m_1)+2m_1u_1}{m_1+m_2} \\ \\ \Rightarrow v_2= \frac{-5(4-6)+2 \times 6 \times 10}{4+6} \\ \\ \Rightarrow v_2= \frac{10+120}{10} \\ \\ \Rightarrow v_2= 13\ m/s$

Answer 2

Velocity of second body after collision (v₂) is -1 m/s.

Given,

Mass of body₁ = m₁ = 6 kg

Velocity of body₁ = u₁ = 10 m/s

Mass of body₂ = m₂ = 4 kg

Velocity of body₂ = u₂ = 5 m/s

The bodies collide elastically

To Find,

Velocity of second body after collision (v₂)

Solution,

Since 2nd body is travelling in opposite direction -

u₂ = - 5 m/s

Let the speed of the body₁ and body₂ after the collision be v₁ and v₂ respectively.

We know, that the coefficient of restitution (e) = 1 for elastic collision

e = $\frac{Relative velocity after collision}{Relative velocity before collision}$

1 = $\frac{Relative velocity after collision}{Relative velocity before collision}$

Relative velocity before collision = Relative velocity after collision

u₁ - u₂ = v₁ - v₂

10 - (- 5) = v₁ - v₂

15 = v₁ - v₂         -   (1)

Now, applying conservation of momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

6(10) + 4(5) = 6v₁ + 4v₂

60 + 20 = 6v₁ + 4v₂

80 = 6v₁ + 4v₂   -  (2)      

Solving (1) and (2), we get,

v₁ = 14 m/s

v₂ = - 1 m/s

Thus, velocity of second body is - 1 m/s.

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