A particle of mass 1 kg is projected upwards with velocity 60 m/s. At the same instant another particle of mass 2 kg is dropped from a certain height. After 2 s (g = 10 m/s2)
(a) magnitude of acceleration of COM = 10 m/s2
(b) velocity of COM = zero
(c) displacement of COM = 20 m
(d) displacement of COM = 40 m.
Answers
it is given that a particle of mass , m = 1kg is projected upwards with velocity, u = 60 m/s at the same instant another particle of mass , M = 2kg is dropped from a certain height.
after 2 sec, velocity of first particle, v₁ = u + at
= 60 - 10 × 2 = 40 m/s.
displacement of first particle, s₁ = 60 × 2 - 1/2 × 10 × 4 = 120 - 20 = 100m
velocity of 2nd particle, v₂ = u + at = 0 - 10 × 2 = -20 m/s
displacement of 2nd particle, s₂ = -1/2 × 10 × 2² = -20 m
now acceleration of centre of mass = (m₁a₁ + m₂a₂)/(m₁ + m₂)
= (1kg × -10 + 2kg × -10)/(1 + 2)
= -10 m/s²
hence, magnitude of acceleration of COM = 10m/s²
velocity of centre of mass, v = (m₁v₁ + m₂v₂)/(m₁ + m₂)
= (1kg × 40m/s + 2kg × -20)/(1kg + 2kg)
= 0 m/s
displacement of COM = (m₁s₁ + m₂s₂)/(m₁ + m₂)
= (1kg × 100m + 2kg × -20m)/(1kg + 2kg)
= 20m
therefore, options (a), (b) and (c) are correct choices.
Answer:
after 2 sec, velocity of first particle, v₁ = u + at
= 60 - 10 × 2 = 40 m/s.
displacement of first particle, s₁ = 60 × 2 - 1/2 × 10 × 4 = 120 - 20 = 100m
velocity of 2nd particle, v₂ = u + at = 0 - 10 × 2 = -20 m/s
displacement of 2nd particle, s₂ = -1/2 × 10 × 2² = -20 m
now acceleration of centre of mass = (m₁a₁ + m₂a₂)/(m₁ + m₂)
= (1kg × -10 + 2kg × -10)/(1 + 2)
= -10 m/s²
hence, magnitude of acceleration of COM = 10m/s²
velocity of centre of mass, v = (m₁v₁ + m₂v₂)/(m₁ + m₂)
= (1kg × 40m/s + 2kg × -20)/(1kg + 2kg)
= 0 m/s
displacement of COM = (m₁s₁ + m₂s₂)/(m₁ + m₂)
= (1kg × 100m + 2kg × -20m)/(1kg + 2kg)
= 20m