Laws of Motion

Question 1

There are two inclined surfaces of equal length (L) and same angle of inclination 45° with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction (μk) between the object and the rough surface is close to

Accepted Answer

0.75. Starting from rest over the same length, t is proportional to 1/√a, so taking twice as long means one quarter the acceleration. With a_smooth = g sinθ and a_rough = g(sinθ − μk cosθ), setting sinθ = 4(sinθ − μk cosθ) gives μk = (3/4) tanθ, which is 0.75 at 45°.

Answer

0.25

Answer

0.40

Answer

0.5

Question 2

A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take g = 9.8 m/s²)

Accepted Answer

21 N s. Impulse equals the change in momentum, and momentum is a vector. The ball arrives at √(2 x 9.8 x 40) = 28 m/s downward and leaves at √(2 x 9.8 x 10) = 14 m/s upward, so J = 0.5 x (14 − (−28)) = 21 N s.

Answer

7 N s

Answer

0

Answer

84 N s

Question 3

A car travels on a circular racetrack of radius 50 m, which is banked at an angle θ. If the car travels at a speed 10 m s⁻¹, then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be 10 m s⁻², the value of θ is:

Accepted Answer

tan⁻¹(1/5). Wear is least when the tyres need no sideways friction at all, which happens at the design speed where the banking alone supplies the centripetal force: tan θ = v²/(rg). Substituting v = 10 m s⁻¹, r = 50 m and g = 10 m s⁻² gives tan θ = 100/500 = 1/5.

Answer

tan⁻¹(2/5)

Answer

tan⁻¹(√3/2)

Answer

tan⁻¹(2√3)

Q2.A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take g = 9.8 m/s²)

Q3.A car travels on a circular racetrack of radius 50 m, which is banked at an angle θ. If the car travels at a speed 10 m s⁻¹, then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be 10 m s⁻², the value of θ is: