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.
Answers
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°
mass of ball a , = 20kg
initial velocity of ball a, = 2m/s i
mass of ball b , = 10kg
initial velocity of ball, b = 0
after collision, ball a moves off with speed of 1m/s at an angle 30° with its original direction.
e.g.,
we have to determine final velocity of ball b
use law of conservation of linear momentum,
e.g.,
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