Physics

Example : 8
A bullet of mass 20 gm moving with a velocity 200 m/s gets embedded into a wooden block of mass 980 gm suspended by a string. Calculate velocity acquired by the combined system.
Solution. Mass of the bullet, m₁ = 20 gm
kg = 20/1000 = 0.02 kg
Velocity of the bullet, u₁ = 200 m/s
Momentum of the bullet = m₁u₁
= 0.02 × 200 kg-m/s = 4 kg-m/s
Now, the bullet gets embedded into a wooden block of mass 980 gm. The mass of the block and bullet = 980 + 20 = 1000 gm = 1kg
Let the velocity of the combined system = v
∴ Momentum of the combined system
= 1 × v kg-m/s = v kg-m/s
Now, applying the law of conservation of momentum,
m₁u₁ = (m₁ + m₂)v
or 4 = v
∴ v = 4 m/s = 4 kg m/s

Example : 9
A rifle man, who together with his rifle has a mass of 100 kg, stands on a smooth surface fires 10 shots horizontally. Each bullet has a mass 10 gm a muzzle velocity of 800 m/s. What velocity does rifle man acquire at the end of 10 shots?
SolLet m₁ and m₂ be the masses of bullet and the rifleman and v₁ and v₂ their respective velocities after the first shot. Initially the rifleman and bullet are at rest, therefore initial momentum of system = 0.
As external force is zero, momentum of system is constant
i.e. initial momentum = final momentum
= m₁v₁ + m₂v₂
or = – 0.08 m/s
Velocity acquired after 10 shots
= 10 v₂ = 10 x (–0.08) = – 0.8 m/s.
i.e, the velocity of rifle man is 0.8 m/s in a direction opposite to that of bullet.

Example : 10
A body of mass 1 kg strikes elastically with another body at rest and continues to move in the same direction with one fourth of the initial velocity. What will be the mass of the other body ?
Sol.Given that,
Initial velocity = u
Final velocity = u/4
So by conservation of momentum, we have
1 x u + 0 = 1 × u/4 + m x v₂
mv₂ = 3u/4 ...(1)
and by conservation of energy, we have

ormv₂² = 15/16 u² ...(2)
from equation (1) and (2),
=
or m = 0.6 kg