Unit 2: Kinematics — Online MCQ Test
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Practice Unit 2: Kinematics with a free chapter-wise online MCQ test.
This chapter covers: The frame of reference motion in a straight line Position- time graph speed and velocity; Uniform and non-uniform motion average speed and instantaneous velocity uniformly accelera....
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Unit 2: Kinematics — Important Questions & Answers
A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is: (g= 10 m/s²)
- A. 340 m
- B. 320 m
- C. 300 m
- D. 360 m
Answer: C. 300 m
Using v² = u² + 2gh, where v=80, u=20, g=10. 80² = 20² + 2(10)h => 6400 = 400 + 20h => 6000 = 20h => h = 300 m.
Using v² = u² + 2gh, where v=80, u=20, g=10. 80² = 20² + 2(10)h => 6400 = 400 + 20h => 6000 = 20h => h = 300 m.
There are two inclined surfaces of equal length (L) and same angle of inclination 45º 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
- A. 0.25
- B. 0.40
- C. 0.5
- D. 0.75
Answer: D. 0.75
Time t = sqrt(2L/a). t_rough = 2 * t_smooth. So acceleration a_smooth = 4 * a_rough. a_smooth = g sin 45, a_rough = g(sin 45 - μ cos 45). Solving g/sqrt(2) = 4 * g(1 - μ)/sqrt(2) leads to μ = 0.75.
Time t = sqrt(2L/a). t_rough = 2 * t_smooth. So acceleration a_smooth = 4 * a_rough. a_smooth = g sin 45, a_rough = g(sin 45 - μ cos 45). Solving g/sqrt(2) = 4 * g(1 - μ)/sqrt(2) leads to μ = 0.75.
In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = x^2 + x. The acceleration of the particle is
- A. -3/2(x+2)^-3
- B. -3/2(2x+1)^-3
- C. 3/2(x+1)^-3
- D. 2/21+x
Answer: B. -3/2(2x+1)^-3
dt/dx = 2x + 1. v = dx/dt = 1/(2x+1). a = dv/dt = (dv/dx)*(dx/dt) = d/dx(2x+1)^-1 * (2x+1)^-1 = -1(2x+1)^-2 * 2 * (2x+1)^-1 = -2/(2x+1)^3.
dt/dx = 2x + 1. v = dx/dt = 1/(2x+1). a = dv/dt = (dv/dx)*(dx/dt) = d/dx(2x+1)^-1 * (2x+1)^-1 = -1(2x+1)^-2 * 2 * (2x+1)^-1 = -2/(2x+1)^3.
Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl driving at 60 km/h from X to Y notices a bus passing her every 30 minutes in her direction and every 10 minutes in the opposite. Find T and bus speed v.
- A. 9 min, 40 km/h
- B. 25 min, 100 km/h
- C. 10 min, 90 km/h
- D. 15 min, 120 km/h
Answer: D. 15 min, 120 km/h
Relative speed in same direction: v-60 = d/30. Relative speed in opposite: v+60 = d/10. Solving for v and T (where T=d/v) yields 15 min and 120 km/h.
Relative speed in same direction: v-60 = d/30. Relative speed in opposite: v+60 = d/10. Solving for v and T (where T=d/v) yields 15 min and 120 km/h.
If the velocity of a particle is v = At + Bt^2, where A and B are constants, then the distance travelled by it between 1s and 2s is:
- A. 3A + 7B
- B. 3/2A + 7/3B
- C. A/2 + B/3
- D. 3/2A + 4B
Answer: B. 3/2A + 7/3B
Distance x = ∫v dt from 1 to 2. ∫(At + Bt^2)dt = A(t^2/2) + B(t^3/3). Evaluating from 1 to 2 gives [A(4/2) + B(8/3)] - [A(1/2) + B(1/3)] = A(3/2) + B(7/3).
Distance x = ∫v dt from 1 to 2. ∫(At + Bt^2)dt = A(t^2/2) + B(t^3/3). Evaluating from 1 to 2 gives [A(4/2) + B(8/3)] - [A(1/2) + B(1/3)] = A(3/2) + B(7/3).
A particle moves so that its position vector is given by r = (cos ωt)i + (sin ωt)j. Where ω is a constant. Which of the following is true?
- A. Velocity and acceleration both are parallel to r
- B. Velocity is perpendicular to r and acceleration is directed towards the origin.
- C. Velocity is perpendicular to r and acceleration is directed away from the origin.
- D. Velocity and acceleration both are perpendicular to r
Answer: B. Velocity is perpendicular to r and acceleration is directed towards the origin.
Given r = (cos ωt)i + (sin ωt)j. Velocity v = dr/dt = -(ω sin ωt)i + (ω cos ωt)j. Acceleration a = dv/dt = -(ω^2 cos ωt)i - (ω^2 sin ωt)j = -ω^2 r. Since a is a negative multiple of r, it is directed towards the origin. Also, r.v = 0, so velocity is perpendicular to r.
Given r = (cos ωt)i + (sin ωt)j. Velocity v = dr/dt = -(ω sin ωt)i + (ω cos ωt)j. Acceleration a = dv/dt = -(ω^2 cos ωt)i - (ω^2 sin ωt)j = -ω^2 r. Since a is a negative multiple of r, it is directed towards the origin. Also, r.v = 0, so velocity is perpendicular to r.
A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is v in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel respectively)?
- A. Both the points P and Q move with equal speed.
- B. Point P has zero speed.
- C. Point P moves slower than point Q.
- D. Point P moves faster than point Q.
Answer: D. Point P moves faster than point Q.
For pure rolling, the topmost point has speed 2v relative to ground and the lowest point has zero speed. Hence point P moves faster than point Q.
For pure rolling, the topmost point has speed 2v relative to ground and the lowest point has zero speed. Hence point P moves faster than point Q.
A particle moving with uniform speed in a circular path maintains:
- A. constant velocity but varying acceleration.
- B. varying velocity and varying acceleration.
- C. constant velocity.
- D. constant acceleration.
Answer: B. varying velocity and varying acceleration.
In uniform circular motion, speed is constant but velocity changes continuously in direction, so velocity is varying. Acceleration also changes direction, so it is varying as well.
In uniform circular motion, speed is constant but velocity changes continuously in direction, so velocity is varying. Acceleration also changes direction, so it is varying as well.
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t1. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t2. The time taken by her to walk up on the moving escalator will be
- A. (t2 - t1) / (t1 * t2)
- B. (t1 * t2) / (t1 + t2)
- C. t1 - t2
- D. t1 + t2
Answer: B. (t1 * t2) / (t1 + t2)
Velocity of girl v1 = d/t1, velocity of escalator v2 = d/t2. Combined velocity v = d/t1 + d/t2 = d(1/t1 + 1/t2). So 1/t = 1/t1 + 1/t2 => t = (t1*t2)/(t1+t2).
Velocity of girl v1 = d/t1, velocity of escalator v2 = d/t2. Combined velocity v = d/t1 + d/t2 = d(1/t1 + 1/t2). So 1/t = 1/t1 + 1/t2 => t = (t1*t2)/(t1+t2).
A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of 80 m/s. The height of the tower is: (g = 10 m/s²)
- A. 320 m
- B. 300 m
- C. 360 m
- D. 340 m
Answer: B. 300 m
Using v² = u² + 2gh: (80)² = (20)² + 2×10×h → 6400 = 400 + 20h → 20h = 6000 → h = 300 m.
Using v² = u² + 2gh: (80)² = (20)² + 2×10×h → 6400 = 400 + 20h → 20h = 6000 → h = 300 m.