Question

A steel ball A of mass 20.0 kg moving with a speed of 2.0 ms−1 collides with another ball B of mass 10.0 kg which is initially at rest. After the collision A moves off with a speed of 1.0 ms−1 at an angle of 30º with its original direction of motion. Determine thefinal velocity of B.​

Answer 1

mass of ball a , $m_a$ = 20kg

initial velocity of ball a, $u_a$= 2m/s i

mass of ball b , $m_b$= 10kg

initial velocity of ball, b $u_b$= 0

after collision, ball a moves off with speed of 1m/s at an angle 30° with its original direction.

e.g., $v_a=1cos30^{\circ}\hat{i}+1sin30^{\circ}\hat{j}$

we have to determine final velocity of ball b.

use law of conservation of linear momentum,

e.g.,$m_au_a+m_bu_b=m_av_a+m_bv_b$

or, 20 × 2i + 10 × 0 = 20 × (1 cos30° i + 1sin30° j) + 10 × v

or, 40i = 20(√3/2 i + 1/2 j) + 10v

or, (40 - 10√3)i - 20j = 10v

or, (4 - √3)i - 2j = v

hence, magnitude of final velocity 3.02 m/s ≈ 3m/s