KPK G11 Physics Chapter 3 Notes (Motion and Force)
Conceptual Questions (Motion and Force)
Table of Contents
Q.1) If you are riding on a train that speeds past another train moving in the same direction on an adjacent track, it appears that the other train is moving backward. Why?
Answer: Velocity is relative. The velocity of an object measured by an observer depends on the motion of the observer itself. If train 1 is moving with a velocity v1 , train 2 is moving with a velocity v2 and both are moving along the same direction , then the velocity of the train 2 with respect to train 1 is: v21 = v2 − v1 If v1 > v2, the above expression becomes negative. This is the reason why the slower train appears to be moving backward.
Q.2) Can the velocity of a body reverse the direction when acceleration is constant? If you think so, give an example.
Answer: Yes, the velocity of a body can reverse its direction with constant acceleration. When we throw a ball vertically upward it will move under constant gravitational acceleration in a direction along positive y-axis. After attaining maximum height it will start falling under same gravitational acceleration whose values is 9.8 ms-2. During its fall its velocity is along negative y-axis. So in this case ball reverses the direction of its velocity while having a constant acceleration.
Q.3) When you stand still on the ground, how large a force does the ground exert on you? Why doesn’t this force make you rise into the air?
Answer: When we are standing on the ground, we are exerting a force equal to our weight due to gravity on the ground and it is pulling us down. On the other hand the ground is reacting by pushing us back; we call this force the reaction force (Newton’s Third law). Hence the ground exert a reaction force on us equal in magnitude to our weight but opposite in direction i.e. both the forces cancels each other and we are not risen in the air.
Q.4) A man standing on the top of a tower throws a ball vertically up with certain velocity. He also throws another ball vertically down with the same speed. Neglecting air resistance, which ball will hit the ground with higher speed?
Answer: Both balls will strike the ground with same speed. As balls are given same initial speed by the man but one is thrown upwards and the other is thrown downwards. Both balls will move with same speed as both are moving under the action of gravity which is constant for all bodies. The only difference is that they both will strike the ground at different times due to different heights. The second ball will take less time compared to first one, while both balls will hit the ground with the same speed as both of them are experiencing same gravitational force and hence experience a constant acceleration with the magnitude of 9.8 ms-2.
Q.5) The cricket coach explains that the follow-through with the shot will make the ball travel a greater distance. Explain the reasoning in terms of the impulse-momentum theorem.
Answer: During the follow through, the bat is in contact with the ball for a longer time. Following through is letting that energy dissipate slowly rather than trying to force it to stop at the point of contact. According to impulse – momentum theorem, the change in momentum is equal to the impulse. In order to impart more momentum into the ball so that the ball travels a greater distance, a greater impulse is needed which is achieved by a longer time of contact between bat and ball.
Answer: When we release an inflated but untied balloon which is initially at rest, the gas inside the balloon rushes out the open end of the balloon. The gas which escapes from the balloon have some momentum. According to law of conservation of momentum the total momentum of a system remains conserved. So, in order to conserve the momentum the balloon acquires a momentum exactly opposite to the momentum of the escaping gas and hence it moves in the direction opposite to that of escaping gas.
Q.7) Modern cars are not rigid but are designed to have ‘crumple zones’ (irregular fold) that collapse upon impact. What is the advantage of this new design?
Answer: A crumple zone is an area of a vehicle (usually located in the front and rear) that is designed to crumple or crush when hit with significant force. The crumple zone helps cushion the initial blow of a collision by increasing the deceleration time so the car stops (relatively) slowly, rather than stopping suddenly which minimizes the force before it reaches the vehicle’s occupants.
Q.8) Why we can hit a long sixer in a cricket match rather than if we toss a ball for ourselves?
Answer: When we toss the ball ourselves there is no pace(speed of ball) to work with, how far the ball travels depends solely on how hard the bat is swung (bat speed) .
Total K.E = bat speed
Now during a match when a batsman hits a ball there are two factors; bat speed as well as the ball speed.
Total K.E = bat speed +ball speed
Their relative speeds add up on contact which results in larger kinetic energy, so the ball can travel farther.
Q.9) An aeroplane while travelling horizontally, dropped a bomb when it was exactly above the target, the bomb missed the target. Explain.
Answer: During the fall, the bomb has both horizontal components of velocity and the vertical component of velocity. When the bomb reaches the target it falls slightly ahead of the target due to the horizontal component of velocity. That is the reason that aeroplane drops the bomb exactly above the target but missed it. So, in order to make it fall exactly on the target, the aeroplane has to drop it slightly before the target.
Q.10) Calculate the angle of projection for which kinetic energy at the summit is equal to one-fourth of its kinetic energy at point of projection.
Answer: Let’s say an object of mass ‘m’ is projected at an angle ‘θ’ with horizontal axis. Therefore, its initial KE
𝐾. 𝐸𝑖 = 1/2 𝑚𝑣2
Since the projection is at an angle with the horizontal, so the velocity of the object v is split into its parallel and perpendicular components;
𝑣𝑥=𝑣𝑐𝑜𝑠𝜃 & 𝑣𝑦=𝑣𝑠𝑖𝑛𝜃
At the highest point of its trajectory, the velocity 𝑣𝑦=0. Therefore,
𝐾. 𝐸𝑓 = 1/2 𝑚𝑣𝑥2 = 1/2 𝑚𝑣2𝑐𝑜𝑠2𝜃
Since the kinetic energy at the highest point of its trajectory is equal to one fourth of its kinetic energy at point of projection, So
𝐾. 𝐸𝑓 = 1/4 𝐾. 𝐸𝑖 1/2 𝑚𝑣2 𝑐𝑜𝑠2𝜃 = 1/4 ( 1/2 𝑚𝑣2 )
Thus 𝑐𝑜𝑠2𝜃 = 1/4
Taking square root on both sides:
𝑐𝑜𝑠𝜃 = 1/2 So θ = 60°
Comprehensive Questions Class 11 Physics Notes for kpk
Q.1) Explain displacement-time graph and velocity-time graph. In each type give brief details along with appropriate diagram for illustration.
Answer: Displacement-Time Graph: With displacement-time graph we can calculate the velocity of body. Displacement-time graph of a moving car is shown in fig (a). From the slope of this graph the velocity of car can be calculated as;
Q.2) Apply Newton’s Laws to explain the motion of objects in a variety of context.
Answer: Newton’s Laws of Motion: Sir Isaac Newton was one of the greatest scientists and mathematicians. He had new different ideas about motion, which he described in his three laws known as Newton’s laws of motion.
Newton’s First Law of Motion: This law states that; An object remains at rest, or in uniform motion in a straight line, unless it is compelled to change by an external net (resultant) force.
Mathematical Expression: ∑F = Fext = 0 Then,Acceleration = a = 0 Explanation: First law of motion is also known as the law of inertia, where inertia is the property of a body to resist the change. When we apply some force on a body to change its state, then the body resists that change. We can say that all bodies try to remain at their stable state and they resist to change. And as Newton’s first law of motion defines the property of inertia of a body so we also name it as the law of inertia. Example: When a cyclist applies sudden breaks, his body feels a thrust in forward direction. This is because when he suddenly applied the brakes his lower part of the body also comes to at rest suddenly, as it was in contact with the bicycle. But his upper portion of the body which was still in the state of motion tried to maintain its state due to inertia. So that is the reason that he felt a push in forward direction.
Newton’s Second Law of Motion: Second law of motion states; A net force applied on the body produces acceleration in the body which is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object.
By combining, we get
From this, we get F = ma Above equation represents the mathematical form of Newton’s second law of motion. Example: When a cricketer catches a ball, he moves his hands in backward direction. This make the acceleration of ball to be zero slowly and not at sudden. By doing so, he save himself from a sudden push from ball. This makes him catch the ball easily without any pain.\
Newton’s Third Law of Motion: This law states; When one object exerts a force on a second object, the second object exerts a force of the same magnitude and opposite direction on the first object.
Q.3) What is linear momentum? Derive and state Newton’s second law in terms of linear momentum.
Answer: Linear Momentum: “The product of mass and velocity is known as linear momentum”
Mathematical Expression: Mathematically, it is given by
P = m v
And in vector form it is,
Unit of linear momentum is kg ms-1 and it can also expressed as N s.
Newton’s Second Law and Linear Momentum: Newton’s second law of motion can be expressed in form of linear momentum. As Newton’s second law of motion states,
By definition of linear momentum;
Substituting value of P in the Newton’s 2nd law relation gives;
This equation states that rate of change of linear momentum of a body is equal to the force acting on the body.
Q.4) State and explain law of conservation of linear momentum for an isolated system of bodies.
Answer: Law of Conservation of Linear Momentum: This law states that, “If there is no external force applied to a system, the linear momentum of that system remains constant in time”
i.e. ΔP = 0
For an isolated system there is no net force acting i.e. F = 0, therefore Newton’s second law in terms of momentum can be written as;
F = ΔP / Δt
As F = 0 so,
ΔP / Δt = 0 Pf– Pi / Δt = 0
By cross multiplication;
Pf – Pi= 0
Pf = Pi
or ΔP = 0
i.e. in the absence of external force the final momentum Pf of the system must be equal to initial momentum Pi i.e. the total momentum of the system can not change.