A bullet of mass 10 g is horizontally fired with velocity 300 m s⁻¹ from a pistol of mass 1 kg. What is the recoil velocity of the pistol?
- (a)0·3 m s⁻¹
- (b)3 m s⁻¹
- (c)–3 m s⁻¹
- (d)–0·3 m s⁻¹
Correct — C, –3 m s⁻¹. Pistol and bullet together form an isolated system in the horizontal direction, so the total momentum before firing, which is zero, must equal the total momentum after. Converting the units first, the bullet's mass is 10 g = 0·01 kg. Then 0·01 × 300 + 1 × V = 0, which gives V = −3 / 1 = −3 m s⁻¹. The magnitude of 3 metres per second says the pistol moves a hundred times slower than the bullet because it is a hundred times heavier, and the minus sign says it moves in the opposite direction to the bullet. Since the options offer both signs, the sign is part of the answer and cannot be dropped.
- (a)0·3 m s⁻¹ — This comes from a unit slip — leaving the bullet's mass at 10 instead of converting 10 g to 0·01 kg, or dividing by 10 somewhere in the arithmetic. It is also unsigned, so it fails on direction as well as on magnitude.
- (b)3 m s⁻¹ — The magnitude is right but the sign is missing. When an option set deliberately includes both +3 and −3, it is testing whether you noticed that recoil is in the opposite direction to the bullet, so the negative value is the one required.
- (d)–0·3 m s⁻¹ — The direction is right but the magnitude is wrong by a factor of ten, again from mishandling the gram-to-kilogram conversion. Checking that 0·01 × 300 gives 3 kg m s⁻¹ of momentum settles it.
The law of conservation of linear momentum follows from Newton's third law: when no external force acts on a system, its total momentum stays constant. In a firing problem the system starts at rest, so the momenta of the two pieces afterwards must be equal in magnitude and opposite in direction, m₁v₁ = −m₂v₂. Because the products are equal, the heavier body always ends up with the smaller speed — the ratio of the speeds is the inverse of the ratio of the masses. This is the same principle behind a rocket's thrust, the backward push felt when jumping off a boat, and the movement of a person standing on frictionless ice who throws an object away.
Two things decide this question, and neither of them is the physics. The first is the unit conversion, since the masses are given in different units and the bullet's must be brought to kilograms before anything else. The second is the sign, and the option set has been built specifically to test it — the same number appears twice, once positive and once negative. Recoil is defined relative to the bullet's direction, so if the bullet is taken as positive the pistol must be negative. A quick sanity check is available too: the pistol is a hundred times heavier than the bullet, so its speed should be a hundredth of 300, that is 3, and any answer of 0·3 or 30 signals an arithmetic slip.
- For an isolated system, total linear momentum is conserved: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
- A gun and bullet start at rest, so their final momenta are equal in magnitude and opposite in direction.
- The recoil speed equals (mass of bullet × velocity of bullet) ÷ mass of gun.
- The heavier body always recoils with the smaller speed, in inverse proportion to the masses.
- Momentum is a vector, measured in kg m s⁻¹, so its direction must be carried through the calculation.
The magnitude comes from the mass ratio and the minus sign from the direction — together they give option (c).
- Forgetting to convert grams to kilograms before applying the momentum equation.
- Reporting only the magnitude when the option set distinguishes the sign.
- Confusing the conservation of momentum with the conservation of kinetic energy, which does not hold here.
As a one-step recoil numerical, or as an assertion-and-reason item on why an isolated system's momentum cannot change.
Assertion (A): A man standing on a completely frictionless surface can propel himself by whistling. Reason (R): If no external force acts on a system, its momentum cannot change.
- (a) Both A and R are true, and R is the correct explanation of A
- (b) Both A and R are true, but R is not a correct explanation of A
- (c) A is true, but R is false
- (d) A is false, but R is true
Answer(a) Both A and R are true, and R is the correct explanation of A
The same principle stated as a law rather than a sum — expelling air one way pushes the body the other, exactly as the bullet pushes the pistol back.
When a ball bounces off the ground, which of the following changes suddenly? (Assume no loss of energy to the floor)
- (a) Its speed
- (b) Its momentum
- (c) Its kinetic energy
- (d) Its potential energy
Answer(b) Its momentum
Drills the point that decides this question — momentum is a vector, so its sign and direction matter even when the magnitude looks unchanged.
- practice — not a real PYQ
A bullet of mass 20 g is fired with a velocity of 200 m s⁻¹ from a rifle of mass 4 kg. The recoil speed of the rifle is
- (a)0·5 m s⁻¹
- (b)1 m s⁻¹
- (c)2 m s⁻¹
- (d)4 m s⁻¹
Answer(b) 1 m s⁻¹ — the bullet carries 0·02 × 200 = 4 kg m s⁻¹ of momentum, so the 4 kg rifle must move backwards at 1 m s⁻¹.
- practice — not a real PYQ
Two skaters of unequal mass push off from each other on frictionless ice. Immediately afterwards
- (a)both move with the same speed
- (b)the heavier skater moves faster
- (c)the lighter skater moves faster
- (d)both remain at rest
Answer(c) the lighter skater moves faster — their momenta are equal and opposite, so speed is inversely proportional to mass.