Question

Ball A of mass 2 kg collides head-on with
ball B of mass 4 kg. A is moving in + x
direction with speed 50 m s–1 and B is
moving in –x direction with speed 40 m
s–1. What are the velocities of A and B
after collision? The collision is elastic.

Answer 1

Answer:

The velocities of A and B after the collision are $v_A$ = -70 m/s and $v_B$ = +20 m/s.

Explanation:

From the given statements, we know

Mass of body A = $m_A$ = 2 kg

Mass of body B = $m_B$ = 4 kg

Velocity of A before collision = $u_A$ = 50 m/s  

Velocity of A before collision = $u_B$ = -40 m/s (negative because of opposite direction)

As we know, for elastic collision, velocity of body A after collision,

$\mathrm{v}_{\mathrm{A}}=\frac{2 \mathrm{m}_{\mathrm{B}} \mathrm{u}_{\mathrm{B}}+\left(\mathrm{m}_{\mathrm{A}}-\mathrm{m}_{\mathrm{B}}\right) \mathrm{u}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}}$

Substituting the values in the above equation, we find,

$\Rightarrow \mathrm{v}_{\mathrm{A}}=\frac{2 \times 4 \times(-40)+(2-4) \times 50}{2+4}$

$\Rightarrow \mathrm{v}_{\mathrm{A}}=\frac{-320-100}{6}$

$\therefore \mathrm{v}_{\mathrm{A}}=-70 \ \mathrm{m} / \mathrm{s}$

Similarly, velocity of body B after collision,

$\mathrm{v}_{\mathrm{B}}=\frac{2 \mathrm{m}_{\mathrm{A}} \mathrm{u}_{\mathrm{A}}+\left(\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}\right) \mathrm{u}_{\mathrm{B}}}{\mathrm{m}_{\mathrm{A}}+\mathrm{m}_{\mathrm{B}}}$

Substituting the values in the above equation, we find,

$\Rightarrow v_{B}=\frac{2 \times 2 \times 50+(4-2)(-40)}{6}$

$\Rightarrow v_{B}=\frac{200-80}{6}$

$\therefore v_{B}=+20 \ \mathrm{m} / \mathrm{s}$