IB DP Physics 2.2 – Forces – IB Style Question Bank : SL Paper 1

Question

Two identical boxes are stored in a warehouse as shown in the diagram. Two forces acting on the top box and two forces acting on the bottom box are shown.

                 

Which is a force pair according to Newton’s third law?

A 1 and 2

B 3 and 4

C 2 and 3

D 2 and 4

Answer/Explanation

Ans: C

For every action, there is an equal and opposite reaction.
The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The  size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs – equal and opposite action-reaction force pairs. 

This is contact or Normal force , which act in pair according to Newton’s Third Law of motion.

Ref: https://www.iitianacademy.com/ib-physics-unit-2-mechanics-forces-dynamics-notes/

NORMAL FORCE

It measures how strongly one body presses the other body in contact. It acts normal to the surface of contact.

Question

Two forces act on an object in different directions. The magnitudes of the forces are 18 N and 27 N. The mass of the object is 9.0 kg. What is a possible value for the acceleration of the object?

A 0 m {s}^{-2}

B. 0.5 m {s}^{-2}

C. 2.0 m {s}^{-2}

D. 6.0 m {s}^{-2}

Answer/Explanation

Ans: C

Let us find the range of acceleration to see which option is best fitted. 

Maximum acceleration is when both forces are in same direction.

Net force = 18 N +27 n = 45 N, Hence  \(a=\frac{F}{m} = \frac{45}{9}= 5 ms^{-2}\)

Minimum Force is when both forces are in opposite direction

Net Force = 27 N – 18 N = 9 N , Hence \(a=\frac{F}{m} = \frac{9}{9}= 1\; ms^{-2}\). All other result will be in between these two depends on the angle between these two forces.

Hence Suitable option for the question is C

Question

Which statement applies to an object in translational equilibrium?

A. The object must be stationary.

B. The object must be moving with constant acceleration.

C. The resultant force acting on the object must be zero.

D. There must be no external forces acting on the object.

Answer/Explanation

Markscheme

C

Ref: https://www.iitianacademy.com/ib-physics-unit-2-mechanics-forces-dynamics-notes/

A body is said to be in equilibrium when no net force acts on the body.
i.e.,  
Then   and

An object is said to be in equilibrium when there is no external net force acting on it. When an object is in equilibrium, it does not accelerate. If it had a velocity, the velocity remains constant; if it was at rest, it remains at rest.
If all the forces acting on a particular object add up to zero and have no resultant force, then it’s in translational equilibrium. Examples would be a book resting on a bookshelf, or someone walking at a steady, constant speed.
An object that’s not rotating or doing so at a steady speed, the sum of the torques acting on it equaling zero, is at rotational equilibrium. Some examples of this are a Ferris wheel turning at a constant velocity, two children of equal weight balanced on either side of a seesaw, or the Earth rotating on its axis at a steady speed

Question

A constant horizontal force F is applied to a block Y. Block Y is in contact with a separate block X.

The blocks remain in contact as they accelerate along a horizontal frictionless surface. Y has a greater mass than X. Air resistance is negligible.

Which statement is correct?

A. The force F is equal to the product of the mass of Y and the acceleration of Y.

B. The force that Y exerts on X is less than F.

C. The force that Y exerts on X is less than the force that X exerts on Y.

D. The force that Y exerts on X is equal to F.

Answer/Explanation

Markscheme

B

Acceleration of block pushing on block
Solution:
Because of the direction in which force \( \vec{F_{app}} \) is applied, the two blocks form a rigidly connected system. We can relate the net force on the system to the acceleration of the system with Newton’s second law. Here, once again for the x axis, we can write that law as 
\(F_{app} =(m_A+m_B)a\)
where now we properly apply \((F_{app}\) to the system with total mass\(m_A+m_B\). Solving for a and substituting known values, we find
\(a=\frac{F_{app}}{m_A+m_B}\)
Hence we can see as per figure b \(F-F_{YX}=(m_Y)a\) hence \(F_XY\) is less than \(F\)

Question

A student of mass m is in an elevator which is accelerating downwards at an acceleration a.

What is the reading on the force meter?

A. mg

B. mg − ma

C. mg + ma

D. ma − mg

Answer/Explanation

Markscheme

B

When lift is accelerated downward
In this case from Newton’s second law
mg – N = ma
or N = mg – ma

Question

The graph shows how the net force F that acts on a body varies with the distance x that the body has travelled.

After travelling 6 m, the change in the kinetic energy of the body is

A. 0 J.
B. 20 J.
C. 30 J.
D. 60 J.

Answer/Explanation

Markscheme

C

Ref: https://www.iitianacademy.com/ib-physics-unit-2-work-energy-and-power-notes/

The work done by a variable force in displacing a particle from x1 to x2
= area under force displacement graph

WORK-ENERGY THEOREM

Let a number of forces acting on a body of mass m have a resultant force and by acting over a displacement x (in the direction of ), does work on the body, and there by changing its velocity from u (initial velocity) to v (final velocity). Kinetic energy of the body changes.
So, work done by force on the body is equal to the change in kinetic energy of the body.
This expression is called Work energy (W.E.) theorem.
Hence Change is Kinetic Energy is area under the curve
Now Area can be calculated as
Area = \(=\frac{1}{2} \times 10 \times 4 +\frac{1}{2} \times 10 \times 2=20+10=30 \;\)

Question

A constant force of 12 N is applied for 3.0 s to a body initially at rest. The final velocity of the body is 6.0m s–1. What is the mass of the body?

A. 1.5 kg
B. 6.0 kg
C. 24 kg
D. 36 kg

Answer/Explanation

Markscheme

B

From the question 

\(u=0 \; ms^{-1}\)
\(v= 6 \; ms^{-1}\)
\(t=3s\)
\(F=12 N\)

Use the equation of linear motion,

\(v=u+at\)
\(6=0+a \times 3\)
\(6 =3a\)
\(a = 2\)
Also
\(F=ma\)
or 12 = m (2)
\(m = 6 kg\)