Question

One ball of mass 0.600 kg traveling 9.00 m/s to the right collides head on
elastically with a second ball of mass 0.300 kg traveling 8.00 m/s to the
left. After the collision, what are their velocities after collision?
(-2.33 m/s (2.33 m/s to right) and 14.67 m/s (14.76 m/s to left))​

Answer 1

Given:

• m₁ = 0.6 kg
• u₁ = 9 m/s [right]
• m₂ = 0.3 kg
• u₂ = - 8 m/s [left]

To find :

• Velocities of both the balls after collision (v₁,v₂)

Solution:

As per the formula,

$\implies\sf{e = \dfrac{v_2-v_1}{u_1-u_2}}$

e = 1 for elastic collision

$\implies\sf{u_1-u_2= v_2-v_1}$

Now let's substitute the given values,

$\implies\sf{9 -(-8) = v_2-v_1}$

$\implies\sf{ v_2-v_1 = 17....(1) }$

As no external force is acting on the system we can say that the momentum of the system remains conserved.

By applying the law of conservation of momentum,

$\implies\sf{m_1u_1+ m_2u_2 = m_1v_1 + m_2v_2}$

Now let's substitute the given values,

$\implies\sf{0.6 \times 9 + 0.3 \times -8 = 0.6v_1 + 0.3v_2}$

$\implies\sf{5.4 - 2.4 = 0.6v_1 + 0.3v_2}$

$\implies\sf{3 = 0.6v_1 + 0.3v_2}$

$\implies\sf{2v_1 + v_2= \dfrac{3}{0.3}}$

$\implies\sf{2v_1 + v_2= 10 ....(2)}$

Multiplying eq(1) by 2 we get,

$\implies\sf{ 2v_2-2v_1 = 34 ....(3)}$

Adding eq(2) and (3) we get,

$\implies\sf{ 3v_2 = 44}$

$\implies\boxed{\sf{ v_2 = 14.67 m/s}}$

Now,

$\implies\sf{ 14.67 - v_1 = 17}$

$\implies\boxed{\sf{ v_1 = - 2.33 m/s}}$

Answer :

After collision,

• v₁ = - 2.33 m/s
• v₂ = 14.76 m/s