Physics JC1 2009
Tutorial 4: Dynamics
1
1. A car collided into a wall. Upon contact with the wall, a force acts on the car and brought it to a rest abruptly. Explain what will happen to the passenger if he is not wearing a seat belt.
Solution:
When the car stops suddenly, the passenger who is originally in motion is reluctant to change his motion. Without wearing seat belt, the passenger will not have any net force acting on him. Thus, due to his inertia, he will be continuing to move forward, i.e. thrown into the air.
2.(a)A 5.0 kg block is pulled along a horizontal frictionless floor by a cord that exerts a force F of 12 N.
What is the acceleration of the block? [2.2 m s-2 to the right]
(b)The force Fin (a) is slowly increased. What is the value of the force just before the block is lifted off the floor and the block’s acceleration at this instance.
[120 N, 21 m s-2]
(a)Since the block is moving along the frictionless floor, the acceleration of the block has to be in the horizontal plane. Thus the resultant force is also acting in the direction of the horizontal plane.
(Looking at the FBD, only F cos θ is in the horizontal plane)
m s-2
(b)Just before the block is lifted off the floor, N ≈ 0.
At this instance, ay = 0 (there is no change in vertical displacement yet)
Since N = 0,
m s-2
5.N06/III (part)
A car traveling along a flat road may be considered to have three forces acting on it. These are represented in the figure below.
(a)
(i)Explain why it is necessary to regard forces A and C as resultant forces.
(ii)Force A has a magnitude 8200 N and is at an angle of 28 to the vertical. Force B is horizontal and has a magnitude 1500 N.
Calculate
- the weight and the mass of the car.
- the resultant force on the car.
- the acceleration of the car.
Solution:
(i)The resultant force A is the vector sum of the contact forces such as normal forces and friction acting on the front and back tyres by the ground.
The resultant force C is the vector sum of the gravitational forces acting on the individual components of the car (that is why the centre of gravity is not acting through the geometrical centre of the car).
(ii)Since the car is moving along the flat road, the acceleration of the block has to be in the horizontal plane. Thus the resultant force is also acting in the direction of the horizontal plane.
Hence there is no acceleration in the vertical plane, i.e. no resultant force in the vertical direction.
1. Considering forces in the y-direction
W = 8200 cos 28
= 7200 N
m = 7240.2 / 9.81
= 740 kg
The resultant force is also acting in the horizontal plane.
2. Considering forces in the x-direction
Fn = A sin 28° + (– B)
= 8200 sin 28 – 1500
= 2349.7 N
= 2400 N
3.Fn = ma
2349.7 = 738 a
a = 3.2 m s-2
(iii)For the car, motion is impossible without friction. Discuss what is meant by friction and the direction in which it acts on the car. In your answer, suggest another example where friction is useful.
Solution:
Friction resists the relative motion between the two surfaces in contact, i.e. it acts parallel to the surfaces in contact and in opposite direction to the motion or intended motion.
On the car, as the tyres turn clockwise, friction resists the relative motion between the surface of the tyres and the road. It acts in the forward direction, allowing the car to move forward.
Friction is useful in walking. The friction between the surface of the shoe and the surface of the floor allows the person to move forward.
(b)Describe a situation in which motion is produced without friction being required.
Solution:
The release of an object in the earth’s gravitational field results in the downward motion of the object (under free fall).
(note: the keyword is produced. Thus the motion must involved a resultant force)
- An object starts from rest and slides down a smooth inclined plane 80 cm long in 0.50s. Determine the angle of the incline. [40.7°]
Along the incline (downwards
m s-2
Since the block is moving along the inclined plane, the acceleration of the block has to be in the inclined plane. Thus the resultant force is also acting in the direction of the inclined plane.
(Looking at the FBD, only Wsinθ is in the inclined plane)
4. J98/III/1
A cyclist moves down a road inclined at an angle of 6.8° without pedaling. The total weight of the cyclist and the bicycle is 760 N.
(a)A graph of his velocity v is plotted against time t.
(i) Using the graph, determine the initial acceleration of the cyclist.
(ii)1. Using your answer in (a)(i), calculate the accelerating force acting on the cycle and cyclist time t = 0.
2. Hence, determine the resistive force acting on the cycle and cyclist at time t = 0.
(iii)State the magnitude of the resistive forceacting on the cycle and cyclist at time t=30 s
(iv)Suggest why the total resistive force has changed between time t = 0 and t = 30 s.
Solution:
(a)(i)A velocity-time graph is given. Thus the acceleration at any instant can be computed by looking at the gradient of the graph at that instant concerned.
Initial acceleration = initial gradient of graph
m s-2
(ii)1.Since the acceleration of the cycle and the cyclist is known, the net force acting on them can be computed.
N
2.The net force acting on the cycle and the cyclist consists of the N, W and R0.
Since the cycle and cyclist is moving along the inclined plane, the acceleration of the block has to be in the inclined plane. Thus the resultant force is also acting in the inclined plane.
Looking at the FBD, only W sin θ and R0 is acting in the direction of the inclined plane)
N
(iii) When t = 30 s, velocity is constant, i.e. a = 0 and Fn = 0
N
(iv)At time t = 0 s, the only resistive force acting is (static) friction. When time increases to 30s, the cycle and the cyclist picks up speed (i.e. accelerates due to the resultant force acting on them). This causes the air resistance acting them to increase, since magnitude of air resistance increases with speed. Thus the total resistive forces include both (kinetic) friction and air resistance as time increases from 0 s to 30 s.
(b) The cycle is serviced in order to reduce friction and then the journey down the slope is repeated. State and explain what change, if any, will occur in the maximum velocity of the cycle down the slope.
Solution:
With a reduction in friction between the cycle and the ground, the total resistive force would be less at any given speed.Hence at the original maximum speed, there would still be a resultant force accelerating the cycle and the cyclist, causing their maximum velocity to be increase.
Extension: How would the initial acceleration change?
Answer: Increased.
(c) Having descended the slope, the cyclist travels along a horizontal straight section of the road at a speed 7.0 m s-1. When the brakes are applied, the cyclist takes 3.5 s to come to rest.
(i)Calculate the average force opposing motion during the time that the brakes are applied, assuming the cyclist is not pedaling.
(ii) Comment on whether the brakes are efficient enough to bring the cycle to a halt when on the inclined road. [0.70 m s-2, 54 N, 36 N, 90 N, 160 N]
Solution:
(c)(i) Since the cyclist is traveling along a horizontal road, its weight will not contribute any component to the Fn. Thus the force opposing motion is the resultant force.
(ii)Since the braking force of 160 N is greater than the component of weight, i.e. W sin 6.8°, along the slope downwards, the brakes should be efficient enough to bring the cycle to a halt.
6. A helicopter has blades of diameter 5.0 m is hovering above the ground at a particular instance. Its blades are rotating in such a way that they are pushing air downwards at a speed of 18 m s-1. The density of the surrounding air can be taken as 1.02 kg m-3.Calculate the upward force acting on the blades. [6500 N]
Solution:
The air molecules are being pushed downward by the blades.
Downward force on air molecules by blades
By Newton’s Third Law, an upward force of 6500 N is acting on the blades by the air molecules.
7. A toy rocket consists of a plastic bottle which is partially filled with water. The space above the water contains compressed air.
At one instant during the flight of the rocket, water of density is forced through the nozzle of radius r at speed v relative to the nozzle. Determine in terms of , r and v,
(i)the mass of water ejected per unit time from the nozzle
(ii)the rate of change of momentum of the water
Hence show that the accelerating force F acting on the rocket is given by the expressionF = r2 v2 – mg, where m is the mass of the rocket and its contents at the instant considered.
Solution:
(i)
(ii)
The rate of change of momentum of the water is πr2ρv2. This is equivalent to the resultant force acting on the water molecules downwards by the rocket. Thus the water should be exerting the same magnitude of force on the rocket upwards (according to Newton’s 3rd law).
Thus to find the accelerating force F, all the forces acting on the rocket has to be considered, i.e. the force exerted by water molecules and its own weight.
Taking upward to be positive,
8. N91/I/5
When a man is standing in an ascending lift, the magnitude of the force exerted on the man’s feet by the floor is always
Aequal to the magnitude of his weight
Bless than the magnitude of his weight
Cgreater than what it would be in a stationary lift
Dequal to what it would be in a stationary lift
Eequal to the magnitude of the force exerted on the lift floor by his feet
Solution: E
The apparent weight depends on whether the lift is ascending at constant speed or accelerating upwards.
However, the action and reaction forces will always be equal in magnitude (according to Newton’s Third Law).
- NJC/2007/Prelim
Two blocks, X and Y, of masses m and 2m respectively, are accelerated along a smooth horizontal surface by a force F applied to block X, and resisted by force ½ F applied to block Y, as shown in the diagram below.
What is the magnitude of the force exerted by block Y on block X during the acceleration?
Solution:
Since the blocks are moving along the smooth surface, the acceleration of the blocks has to be in the horizontal plane. Thus the resultant force is also acting in the direction of the horizontal plane. (In this case, the normal force and weight of the blocks are not indicated in the FBD)
Since X and Y are moving together, they should have the same velocity and acceleration at any time instant. Let the acceleration of X and Y be a.
For both block X and Y,For Block X
Sub into (2)
10. N2004/III/2
(a) Define
(i) linear momentum
(ii) State the relationship between the change in linear momentum of an object, the constant force acting on the object, and the time for which the force acts.
Solution:
(a(i)The linear momentum (of a body) is the product of its mass and its linear velocity.
(ii)According to Newton’s 2nd law, Fn = dp/dt, thus
The change in linear momentum is the product of force and time interval.
i.e. p = Ft
(b) In a collision between two bodies A and B, the force that A exerts on B varies with time in the way shown in the figure below.
(i)Copy the figure and show on your sketch a graph of the force that B exerts on A.
(ii) Explain your answers to (i)
(iii) Explain how your answers to (i) is consistent with the principle of conservation of momentum.
Solution:
(i)
(ii)According to Newtons’ Third Law, the force exerted on A by B is equal in magnitude but opposite in direction to the force exerted on B by A.
(iii) When considering bodies A and B as a system, the total momentum of A and B remains constant, i.e. no change in the change in momentum. The change in momentum is reflected in terms of the area under the force-time graph.
Since the graphs must be mirror images of one another as any gain in momentum in A must be equal to the loss of momentum of B, resulting in the total momentum of the A and B remaining constant.
(c) In a collision, when a truck of mass 12 000 kg runs into the back of a car of mass 1200 kg, a constant force of 72 000 N acts for 0.25 s.
Calculate the change of velocity of
(i)the car
(ii)the truck [15 m s-1, 1.5 m s-1]
(d)Suggest one way in which the conditions in (c) are unrealistic.
Solution:
(c) (i)pcar = Fcart = mcarv car
72 000 × 0.25 = 1200×v car
vcar = 1.5 m s-1
(ii) Since the car and the truck are interacting with each other, the magnitude of change in momentum of the car and the truck should be the same, in which car will gain momentum while the truck will lose momentum, i.e. the total change in momentum of the car and the truck should be zero (COM)
ptruck = F t = mtruckv truck
- (72 000) (0.25) = (12 000) vtruck
vtruck = - 15 m s-1
(d) The force of impact is not constant. Its magnitude should vary with time as indicated in (b).
(e) Discuss how seat belts and air bags in a car ensure greater safety.
(f) In order to reduce the number of traffic accidents, many countries conduct research into improving road safety.
(i)One area of research concerns braking. State three factors that affect braking which might be considered by researchers.
(ii)State one other aspect of car safety that could be researched, suggest briefly how the research could be conducted.
Solutions:
(e) Seat belts and air bags can provide the backward force to stop the passengers from being flung forward due to inertia.
At the same time, the time of impact (between the body and the safety belts and airbags) is increased, causing the magnitude of the impact force to be reduced with the same change in momentum (p = F t).
(f)(i)- The condition of the road surface. Wet or dry.
- The grid (tread) on the tire.
- The quality of the braking pad.
- Efficiency of braking fluid
(ii)- Crumple area of the car.
- Tyre design
11. N02/II/3 (part)
A steel ball of mass 250 kg is suspended from the jib of a crane. In order to demolish a wall, the ball is pulled away from the wall and then released. The ball swings down and hits the wall.
The variation with time t of the speed v of the ball is shown in Fig 3.2.
(a) Using Fig 3.2, determine
(i) the magnitude of the acceleration of the ball at time t = 0.8 s. [3.1 m s-2]
(ii) the distance moved by the ball before it hits the wall, that is, from time t = 0 s to t = 1.6 s. [4.9 m]
Solution:
(a)(i)A velocity-time graph is given. Thus the acceleration at any instant can be computed by looking at the gradient of the graph at that instant concerned.
Draw a tangent at t = 0.8 s and calculate its gradient.
(1.2, 5.1), (0.15, 2.0)
a = =2.9 m s-1 (within acceptable range)
(ii)To find the distance moved by the ball, the area under the velocity-time graph should be computed. Since it is of irregular shape, the best way is to count the squares.
distance = area under the graph between t = 0 s to t = 1.6 s
= 4.9 m
(b) Calculate the magnitude of
(i)the change in momentum of the ball during its collision with the wall. [1300 kg m s-1]
(ii) the average force exerted on the wall during the collision. [6500 N]
Solution:
(b)(i)p = m v
p = m (vf - vi)
p = 250 (0 - 5.2)
p = - 1300 kg m s-1
(b) (ii) magnitude of average force on wall
= magnitude of average force on ball(According to Newton’s 3rd law)
= p / t
= 1300 / (1.80 – 1.60)
= 6500 N
12. 2008/H1/I/8
A large mass moving at a velocity of 5 m s-1 collides head-on with a small mass moving at a velocity of 2 m s-1 in the opposite direction.
The collision is elastic.
After the collision, both masses move to the right.
The large mass has a velocity v1 and the small mass has a velocity v2.Which pair of values v1 and v2 is possible?
Solution:
In an elastic collision,
relative speed of approach = relative speed of separation.
So the solution is either Option B or D.
Besides that in an elastic collision, the total kinetic energy of the interacting objects remains constant. Since the large mass retains its original velocity while the small mass has its velocity increased from 2 m s-1 to 12 m s-1, therefore total K.E is not conserved in Option D.
This leaves Option B as the answer.
14. N2000/III/1 (part)
(b)In a gas, a hydrogen molecule, mass 2.00 u and velocity 1.88×103 m s-1, collides elastically and head-on with an oxygen molecule, mass 32.0 u and velocity 405 m s-1 as shown below.
In qualitative terms, what can be stated about the subsequent motion as a result of knowing that (i) the collision is elastic,
(ii) the collision is head-on?
(c)Using your answers to (b),
(i)determine the velocity of separation of the two molecules after the collision,
(ii) apply the law of conservation of momentum to the collision,
(iii)determine the velocity of both molecules after the collision.
[2285 m s-1, 136 m s-1, 2420 m s-1]
Solution:
(b) (i)If the collision is elastic, the KE and total momentum of the two objects must be the same as that before the collision.
(ii) If the collision is head-on, the motion of the molecules after collision will be collinear with the motion of the molecules before collision.
(c)(i)
(ii)By conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
(2.0u)(1880) + (32.0u)(-405) = (2.0u) v1 + (32.0u)v2
-4600 = v1 + 16.0v2
(iii)Subst. v1 = - 16.0v2 – 4600 into Eqn (1)
v2 – (– 16.0v2 – 4600) = 2285
v2 = – 136 m s-1(or 136 m s-1 to the left)
Sub back to Eqn (1)
v2 – v1 = 2285
- 136 – v1 = 2285
v1 =– 2420 m s-1(or 2420 m s-1 to the left)
13. 2008/H2/I/6
A trolley of mass 6.0 kg travelling at a speed of 5.0 m s-1 collides head-on and locks together with another trolley of mass 10 kg which is travelling in the opposite direction at a speed of 3.0 m s-1. The collision lasts for 0.20 s.
What is the total momentum of the two trolleys before the collision and the average force acting on each trolley during the collision?
Solution:
Total momentum
= m1 u1 + m2 u2
= 6.0 (5.0) + (10)(-3.0)
= 0 kg m s-1
Since the trolleys lock together after the collision and that the total momentum of the system is 0 kg m s-1, the velocity of the trolleys after collision is 0 m s-1.
Considering the 6.0 kg trolley,
p = F t
0 – (6.0) (5.0) = F (0.20)
F = - 150 N
F = 150 N (in opposite direction to initial velocity)
Answer : C
15. N99/II/1 (part)
(b)(i)A fast-moving neutron of mass m collides head-on with a stationary atom of hydrogen, also of mass m .