Download Normal Reaction force

Year 12 Revision 4.
Name:
There are 9 questions, each is worth one mark unless otherwise stated. Time = 20 minutes.
Jessica moves down a water slide of vertical height 6.0m into a swimming pool as shown in Figure 1 below.
Jessica starts from rest at Q and you may assume that friction and air resistance are negligible.
Question 1
Calculate the speed of Jessica at the point S.
Question 2
Sketch and label the individual forces on Jessica at the point R.
Now consider the forces on Jessica at point S
Question 3
Which one of the arrows( A-F) in Figure2 below best represent the net force on Jessica at point S?
A book is resting on a table as shown in Figure 3 below. N is the force exerted by the table on the book and
mg is the weight of the book.
A class of students is asked to consider the relative magnitude of N and mg. The students are asked to give
their reasoning.
Frank states 'N and mg are equal in magnitude because N is the reaction force to mg in the sense of
Newton's third law'.
Emma states 'N and mg are equal in magnitude because the net force on the book is zero. N is not the
reaction force to mg in the sense of Newton's third law'.
Question 4
2 marks
Explain, giving reasons, which student is correct.
A car is tested on a straight level track on a day when there is no wind.
When the car reaches 20.0 ms-1 the driver put the car into neutral gear. The wheels are no longer driving the
car, and it gradually slows due to frictional forces. Measurements show that the car's speed decreases
uniformly from 20.0ms-1 to 18.0ms-1 in 4.0 s. The mass of the car is 1100 kg.
Question 5
Calculate the magnitude of the net force on the car when it is slowing down.
Question 6
Calculate the power required at the wheels of the car for it to be driven at a constant speed of
20.0m s-1 on the straight level track.
In a tennis serve the ball is hit over the net into the service court. Assume that the ball does not spin.
Air resistance should be taken into account.
Question 7
Sketch and label the individual forces on the ball while it is in mid flight, at the point X in Figure 4 above.
For Questions 8 and 9, neglect the effect of air resistance.
Question 8
Sketch a graph of the ball's horizontal component of velocity as a function of time after the ball leaves the
racquet until just before it hits the court. Numerical values are not required.
In this serve the ball was initially hit horizontally from a height of 3.0m above the surface.
Question 9
Calculate the time of flight from hitting the serve until the ball first hits the court surface on the other side of
the net.
Question 1 Solution

List what data you have and what you need.

Choose an appropriate formula

Be careful with fractions and using your calculator.
The loss in PE is equal to the gain in KE.
So KE =
1
2
mv2 =  mgh = m × 9.8 × 6
2  m  9.8  6
= 117.6
m
 v = 10.8 m/s
 v2 =
Question 2 Solution

We are looking for the forces on Jessica, not on the ramp

Note that there isn’t any friction or air resistance

So, the only two forces acting are weight, (always vertically down) and the normal reaction. Use this
term so that you remind yourself that this force is perpendicular to the surface.

Draw the forces approximately to scale.

Label the forces, this is another command.

Don’t draw in F (net)(unless you are specifically asked for it).
Normal Reaction force
W = mg
Question 3 Solution

At the bottom, the two forces acting are both vertical. the net force will be vertical, and upwards. This
is because the weight is down, and the normal reaction is vertically up.

At this ‘instant’, that is at the bottom, the motion can be modelled as uniform circular motion.
 The net force acting on her must be radially inwards.  A
Question 4 Solution
Emma is correct and Frank is incorrect.
The book is at rest, so, the resultant force on the book is zero. As a result N and mg are equal in magnitude
but opposite in direction.
N and mg both act on the book and therefore cannot be action and reaction in the sense of Newton's 3 rd Law
because action and reaction act on different bodies. In fact, the reaction to N is the force that the book
exerts on the table and the reaction to mg is the gravitational force that the book exerts on the Earth
This question was found to be very hard on the exam, only 8% of the students got the full 2 marks. So make
sure that you understand this concept and this explanation.
Question 5 Solution
v 20  18
=
=0.5 m/s2
4
t
 F = 1100 × 0.5 = 550N
Use F = ma
where a =
Question 6 Solution
For the car to be driven at a constant speed, the sum of the forces must be zero. Therefore the driving force
must be the same magnitude but in the opposite direction to the frictional forces.
Power = F × v = 550 × 20 = 11 000 W
Question 7 Solution
The ball is not spinning, so the air resistance is opposing the motion. The motion at this point will not be
exactly horizontal, but slightly below horizontal.  the air resistance will be to the left and slightly up. The
other force acting on the ball is the weight force. This is acting directly down.
air resistance
weight
Question 8 Solution
If you neglect the effect of air resistance, then the only force acting on the ball is the weight force. Since this
force is directly down, then there aren't any forces acting in the horizontal direction. This means that the ball
will have a constant horizontal velocity.
horizontal velocity
time
Question 9 Solution
0.78 s
The ball needs to fall 3.0 m.
Use x = ut + ½gt2
3.0 = 0 + ½ × 9.8 × t2
 t2 = 3.0  4.9
 t2 = 0.612
 t = 0.78 secs