The motion of a particle of mass m is described by the relation, y = ut − ½ gt², where u is the initial velocity of the particle. The force acting on the particle is
- (a)F = m (du/dt)
- (b)F = mg
- (c)F = m (dy/dt)
- (d)F = − mg
Correct — D, F = − mg. The relation given is the position of the particle, so the force has to be reached by differentiating it twice and then applying Newton's second law. Differentiating y = ut − ½gt² once gives the velocity, dy/dt = u − gt; differentiating again gives the acceleration, d²y/dt² = − g, a constant. For a body of fixed mass, F = m × acceleration, so F = m(− g) = − mg. The minus sign is not decoration — the relation as printed takes the direction of the initial velocity u as positive, and the acceleration comes out opposite to it, so the force acting on the particle points the other way. That force is simply the particle's weight, pulling it back towards the ground.
- (a)F = m (du/dt) — u is the initial velocity — the value fixed at t = 0 — so it is a constant and du/dt is zero. This expression therefore gives F = 0, which cannot be right for a particle that is decelerating and turning round. Newton's second law needs the rate of change of the instantaneous velocity, not of a number set at the start.
- (b)F = mg — The magnitude is right but the sign is wrong. Working from the relation as printed, the acceleration is − g, so the force is − mg; writing + mg would mean the force acts along the direction of launch, which would speed the particle up instead of bringing it back.
- (c)F = m (dy/dt) — dy/dt is the velocity, so m(dy/dt) is mass times velocity — momentum, in kg m s⁻¹. Force is measured in kg m s⁻², so this fails a simple check of units. Force is the rate of change of momentum, which needs one further derivative.
For a body of constant mass, Newton's second law reads F = ma, and acceleration is the second derivative of position with respect to time. So whenever a question hands you a position-time relation and asks for the force, the route is fixed: differentiate twice, then multiply by the mass. The relation y = ut − ½gt² is the standard equation for a particle launched with speed u and then acted on only by a uniform downward pull of magnitude g.
Two things decide this item. The first is knowing which symbol is a variable and which is a constant — t varies, u does not, which kills option (a) at once. The second is respecting the sign convention already built into the printed relation. Because the − ½gt² term carries the minus, the equation is written with the direction of u as positive, and the acceleration that falls out of it is − g rather than + g. A candidate who answers with the magnitude of the weight picks option (b) and loses the mark. Option (c) is worth a moment even if you do not spot the physics, because a units check disposes of it in seconds.
- Differentiating y = ut − ½gt² once gives velocity v = u − gt; differentiating again gives acceleration a = − g.
- Newton's second law for constant mass is F = ma, so a = − g gives F = − mg.
- u is the initial velocity, a constant of the motion, so du/dt = 0.
- m(dy/dt) has the units of momentum (kg m s⁻¹), not of force (kg m s⁻², the newton).
- Answering with the magnitude mg and ignoring the sign the printed relation forces on you.
- Differentiating the initial velocity u, which is a constant, instead of the position y.
- Confusing m(dy/dt), which is momentum, with force — a units check settles it.
Asked as a one-line calculus-and-units check on a printed kinematic relation, with the sign of the answer doing the discriminating.
The mass of a body on Earth is 100 kg (acceleration due to gravity, gₑ = 10 m/s²). If acceleration due to gravity on the Moon = gₑ/6, then the mass of the body on the moon is
- (a) 100/6 kg
- (b) 60 kg
- (c) 100 kg
- (d) 600 kg
Answer(c) 100 kg
Tests the other half of the same relation. Here the mass m stays put while g changes; in the NDA item g is fixed and the question is what m and g together produce, namely a force of magnitude mg.
Weight and mass of an object are defined with Newton's laws of motion. Which among the following is true?
- (a) Weight is a constant of proportionality.
- (b) Mass is a constant of proportionality.
- (c) Mass is not a constant of proportionality.
- (d) Weight is a universal constant.
Answer(b) Mass is a constant of proportionality.
Makes explicit the step this question uses silently — mass is the constant that links force to acceleration in F = ma, which is why multiplying the acceleration − g by m gives the force.
A rigid body of mass 2 kg is dropped from a stationary balloon kept at a height of 50 m from the ground. The speed of the body when it just touches the ground and the total energy when it is dropped from the balloon are respectively (acceleration due to gravity = 9·8 m/s²)
- (a) 980 m s⁻¹ and 980 J
- (b) √980 m s⁻¹ and √980 J
- (c) 980 m s⁻¹ and √980 J
- (d) √980 m s⁻¹ and 980 J
Answer(d) √980 m s⁻¹ and 980 J
The same uniform downward acceleration g, used numerically instead of symbolically — a worked drill on the motion that y = ut − ½gt² describes.
- practice — not a real PYQ
A particle moves so that its position is given by x = at³, where a is a constant. The force acting on the particle is proportional to
- (a)t
- (b)t²
- (c)t³
- (d)a constant, independent of t
Answer(a) t — differentiating twice gives acceleration 6at, so F = m(6at) grows in direct proportion to time.
- practice — not a real PYQ
A ball is thrown vertically upward. While it is rising, the force acting on it (ignoring air resistance) is
- (a)upward and decreasing
- (b)downward and constant
- (c)zero at the highest point
- (d)upward at first and downward later
Answer(b) downward and constant — only the weight acts, so the force is mg downward throughout, including at the highest point where the velocity is zero.