Question

A steel ball a of mass 20.0 kg moving with a speed of 2.0 ms1 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 ms1 at an angle of 30 with its original direction of motion. Determine the final velocity of

b.

Answer 1

Mass of a, m₁=20 kg

Mass of b, m₂=10 kg

Velocity of a before collision u₁= 2m/s

velocity of a after collision u₂=1m/s

Velocity of b after collision v₂= ?

Velocity of b before collision v₁=0

Angle made by a with the original direction α=30°

Let the angle made by b is β

Applying Law of conservation of Momentum along the x- axis

m₁u₁+m₂v₁=m₁u₂Cos30°+m₂v₂Cosβ

20×2+0=20×1×√3/2+10×v₂Cosβ

v₂Cosβ= 4-√3           (1)

Applying Law of conservation of momentum along y-axis

m₁u₁+m₂v₁=m₁u₂Sin30+m₂v₂Sinβ

0+0=20×1×1/2+10v₂Sinβ

v₂Sinβ= -1           (2)

Now 2÷1 , we get,

Tanβ= -1/(4-√3)=336.20°

v₂=(4-√3)/Cos336.20°=2.48m/s

Ans= 2.48m/s along 336.20°

Answer 2

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