A 100 g sphere is moving at a speed of 20 m/s and collides with another sphere of mass 50 g. If the second sphere was at rest prior to the collision and the first sphere comes at rest immediately after the collision, considering the collision to be elastic, the speed of the second sphere would be
- (a)10 m/s
- (b)20 m/s
- (c)30 m/s
- (d)40 m/s
Correct — D, 40 m/s. Work with momentum. Before the collision only the first sphere is moving, so the total momentum of the pair is p = m1 u1 = 0.100 kg × 20 m/s = 2.0 kg m/s. After the collision the first sphere is stated to be at rest, so the whole of that momentum has to be carried by the second sphere alone. Setting m2 v = 2.0 kg m/s with m2 = 0.050 kg gives v = 2.0 / 0.050 = 40 m/s. The lighter body must move faster to carry the same momentum, and it is exactly twice as light, so it ends up exactly twice as fast as the incoming sphere. The answer is 40 m/s.
- (a)10 m/s — This is 20 m/s halved. It comes from multiplying by the mass ratio the wrong way round — treating the lighter sphere as though it must move more slowly. Momentum works the other way round, because less mass carrying the same momentum means more speed, not less.
- (b)20 m/s — This assumes the speed is handed over unchanged, as if velocity were the conserved quantity. It is not. Speed is transferred intact only when the two masses are equal, and here the second sphere is half the mass of the first.
- (c)30 m/s — No consistent reading of the data produces 30 m/s. It sits between the momentum answer of 40 m/s and the roughly 26.7 m/s that the standard elastic-collision formulas would give for these two masses, so it is the number you land on by splitting the difference rather than by applying either rule.
In any collision on which no external force acts, the total linear momentum of the colliding bodies is the same immediately before and immediately after. That single statement is enough to solve most collision problems in this syllabus, because momentum is a product of mass and velocity, so a change in one must be paid for by a change in the other. Kinetic energy is a separate matter — it is conserved only in a perfectly elastic collision and is partly lost as heat, sound and deformation in an inelastic one.
The route to the key's value runs entirely through momentum conservation, and that is the relation the item is really testing. It is worth being straight with students about a tension in the wording. The stem says the collision is elastic and also that the heavier sphere comes to rest, and those two conditions cannot both hold for these masses. In a genuinely elastic head-on collision the incoming body is brought to rest only when the two masses are equal; with 100 g striking 50 g the elastic formulas give about 6.7 m/s for the first sphere and about 26.7 m/s for the second, so the first would not stop. Checking the energy makes the same point — 20 J goes in and the stated outcome would carry 40 J out, which no collision can do. The printed data over-specify the problem. The condition that actually fixes a number, and the one the official key follows, is momentum conservation with the first sphere at rest, and that gives 40 m/s.
- Linear momentum p = mv is conserved in a collision when no external force acts on the colliding system.
- Here the momentum before is 0.100 kg × 20 m/s = 2.0 kg m/s, and after the collision the 50 g sphere must carry all of it.
- Solving 0.050 kg × v = 2.0 kg m/s gives v = 40 m/s.
- A perfectly elastic collision also conserves kinetic energy, and in a head-on elastic collision the incoming body is brought to rest only when the two masses are equal.
The whole 2.0 kg m/s has to leave with the 50 g sphere, and half the mass means double the speed.
- Assuming speed rather than momentum is passed on, which gives 20 m/s.
- Applying the mass ratio upside down and shrinking the speed instead of raising it.
- Reaching for the elastic-collision formulas when the outcome after the collision is already given; momentum conservation alone settles it.
Asked as a one-line numerical item — apply conservation of momentum to a two-body collision in which the final state of one body is given.
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
States in words the principle this question makes you compute with. Momentum of an isolated system cannot change, so air pushed one way must push the man the other — exactly as the momentum lost by the heavier sphere reappears in the lighter one.
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
Answer(c) –3 m s⁻¹
The same one-line calculation in mirror image. There a light body leaves fast and a heavy one recoils slowly; here the momentum passes from heavy to light and the light one speeds up. Both are the single equation total momentum before equals total momentum after.
A railway wagon (open at the top) of mass M₁ is moving with speed v₁ along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M₂ and speed becomes v₂. Taking the rain to be falling vertically and water stationary inside the wagon, the relation between the two speeds v₁ and v₂ is :
- (a) v₁ = v₂
- (b) ½ M₁v₁² < ½ M₂v₂²
- (c) M₁v₁ = M₂v₂
- (d) M₁v₁ < M₂v₂
Answer(c) M₁v₁ = M₂v₂
Asks the same question in symbols and makes the point that momentum, not speed and not kinetic energy, is the quantity that stays put. Once mass changes, the speed must change to compensate — the reasoning behind the 40 m/s here.
- practice — not a real PYQ
A 2 kg trolley moving at 3 m/s collides with a stationary 4 kg trolley and the two move off together. Their common speed is
- (a)0.5 m/s
- (b)1 m/s
- (c)1.5 m/s
- (d)2 m/s
Answer(b) 1 m/s — momentum before is 6 kg m/s, and dividing by the combined 6 kg gives 1 m/s.
- practice — not a real PYQ
In a head-on elastic collision between two bodies, the moving body comes exactly to rest after impact when
- (a)the moving body is much heavier than the stationary one
- (b)the two bodies have equal mass
- (c)the stationary body is much heavier
- (d)it happens for any pair of masses
Answer(b) the two bodies have equal mass — only then can momentum and kinetic energy both be conserved with the first body stopping.