Question

7. a) (i) A bullet of mass 0.02 kg is fired by a gun of mass 20 kg. If the speed of the
bullet is 100 ms-1, calculate the recoil speed of the gun?

(ii) Explain the process of rocket propulsion in the light of Newton’s third law
of motion.

Answer 1

Explanation:

Let the direction in which the bullet is fired be positive.

Therefore,

Mass of Bullet(m)=20 grams=0.02 kg

Mass of Pistol(M)=2 kg

Initial Velocity of Bullet(u)=0 m/s

Initial Velocity of Pistol(U)=0 m/s

Final Velocity of Bullet(v)=150 m/s

Final Velocity of Pistol(V)= Recoil Velocity of Bullet = ?

According to the Law of Conservation of Momentum,

mu + MU = mv + MV

(0.02)(0) + (2)(0) = (0.02)(150) + (2)(V)

0 + 0 = 3 + 2V

-3/2 = V

V = -1.5 m/s

Therefore, the recoil velocity of the pistol will be 1.5 m/s in the opposite direction.(As the sign is negative)

Answer 2

Mass of thee bullet , m = 0.02 kg

Mass of the gun , M = 20 kg

Speed of the bullet , v = 100 m/s

Recoil velocity of gun , V = ? m/s

Apply conservation of momentum ,

$\sf \to\ \; P_i=P_f\\\\\to \sf (m+M)u=mv+MV$

Initially gun and bullet are in rest .

( u = 0 m/s )

$\to \sf (0.02+20)0=(0.02)(100)+(20)V\\\\\to \sf 0=2+20V\\\\\to \sf 20V=-2\\\\\to \sf \pink{V=-0.1\ m/s}\ \; \bigstar$

Note : -ve sign of recoil velocity denotes movement in  opposite direction

i. Recoil velocity = 0.1 m/s

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ii.

Rockets which are mostly used in  NASA works on the principle of Newton 3rds Law . Whenever they produce gas from backward side , rocket itself moves forward direction . Here , action is given from lower side of rocket and reaction is taken by upper part for moving forward .

• Action of gases = Reaction of movement of rocket