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Forces and Motion, Multiple Choice Questions

See here MCQs included in the course of Physics 11.

Choose the best possible answer.

(1) A ball is thrown vertically upwards at 19.6 m/s. For its complete trip (up and back down to the starting position), its average speed is:
A. 19.6 m/sB. 9.8 m/sC. 6.5 m.sD. 4.9 m/s

(2) If you throw a ball downwards, then its acceleration immediately after leaving your hand, assuming no air friction, is:

A. 9.8 m/s2B. more than 9.8 m/s2C. less than 9.8 m/s2D. speed of throw is required for answer
3. The rate of change of momentum gives
A. Force B. Impulse C. Acceleration D. Power

4. The area between velocity-time graph is numerically equal to:  
A. VelocityB. DisplacementC. AccelerationD. Time


Consider the fig to the right. A body is moving with a uniform velocity of 10 m/s for 4 seconds. What is its displacement from the starting position? 10 × 4 = 40 m.

Now calculate the area under the curve. It is

Area = height × width = 10 m/s × 4 s = 40 m

Result: Area under the velocity-time curve is equal to the displacement of the body.


Now take another example. A body starts from rest and reaches a velocity of 10 m/s in 4 s. The graph is as shown in the fig to the right.

Calculate the area under the graph (shaded area). It is a triangle, therefore,

Area = ½ (height × width) = ½ (10 m/s × 4 s) = ½ × 40 m = 20 m

Now imagine about the displacement traveled by the body. As the velocity increases from 0 to 10 m/s uniformly, the average velocity must be ½ (0 + 10) = 5. So the body is moving with an average velocity of 5 m/s. It goes for 4 s, therefore, its displacement is 5 × 4 = 20 m.

Again the result is the area of a velocity-time graph shows the displacement. We can prove in similar way for any type of velocity or combination of increasing, decreasing, uniform etc.

Please note the importance of the word “numerically” in the question. The displacement is equal to the area under the curve in magnitude, however, the directions are different.

5. If the slope of velocity-time graph gradually decreases, then the body is said to be moving with:
(A) Positive acceleration
B. Negative acceleration
(C)Uniform velocity
D. Zero acceleration
Explanation: Mathematically, slope of a graph = change in y-axis ÷ change in x-axis = . Geometrically, slope is the incline in the line. It is the tangent of the angle the graph line makes (perpendicular/base). Physically, it is the acceleration of the moving body (Δv/Δt). Now, time cannot be negative. So, if Δv/Δt is negative, then Δv (change in velocity) must be negative. This means the final value of v is less than the initial value of v and the velocity is decreasing. Decrease in velocity with time is negative acceleration.
  1. A 7.0 kg bowling ball experiences a net force of 5.0 N. What will be its acceleration?

A. 3.5 m/s2

B. 7.0 m/a2

C. 5.0 m/s2

D. 0.71 m/s2

Solution: m = 7.0 kg,       F = 5.0 N,             Formula: F = ma                Put the values

  1. SI unit of impulse is:

A. kg ms-2

B. N s

C. N s-1

D. N m

Hint: Impulse = F.Δt

  1. A ball with original momentum +4.0 kg × m/s hits a wall and bounces straight back without losing any kinetic energy. The change in momentum of the ball is:

A. +4 N s

B. -4 N s

C. +8 N s2

D. -8 N s

Solution: (1) The sign indicates the directions. If going towards the wall is taken as positive direction, then bouncing back is negative.

(2) If no K.E is lost, this means the ball would have numerically the same momentum when bounced back.

(3) Change in momentum, Δp = pf – pi.

Put values, Δp = -4 – (+4) = -4 -4 = -8 N s

(Note: For the unit of momentum, consider p = mv = kg ms-1 = kg ms-1 × s-1× s+1= kg ms-2 × s1 = N s).

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