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Revision for the Motion #2 Unit Test …ANSWERS
1. Explain using Newton’s 1st law, how a skier can ski down slope at constant velocity.
The sum of the forces on an object must be zero if an object continues at constant velocity.
Therefore the component of the force of gravity acting in a direction down the slope must be
balanced by the frictional forces acting upwards against the motion of the skier.
2. Explain using Newton’s 2nd law, how a stone accelerates when it is dropped.
There is an unbalanced force acting on the stone – the force of gravity and since, according to the
second law, F= ma, the gravitational force = proportional to the mass of the stone and equals the
mass of the stone times the acceleration due to gravity. The stone accelerates towards the ground
at a rate given by a = F/m.
----------------------------------------------------------------------------------3. If a 64 kg skier is skiing down a 14o slope at constant speed, determine:(a) the normal reaction force acting on the skier. The fact that the skier is travelling at
constant speed means that the frictional forces acting up the slope are balanced by the
gravitational force acting down the slope. A diagram is essential.
mg
θ
The normal reaction force is the force acting upwards from the slop to balance the component
of the skier’s weight acting down at right angles to the slope.
This equals mgcosθ = 64 x 9.8 x Cos 14 = 609 N
(b)
the friction forces acting on the skier
The frictional forces acting on the skier will be equal and opposite to the component of the skier’s
weight acting down the hill and will equal mgsinθ = 152 N up the slope
----------------------------------------------------------------------------------4. A 1.25 tonne car is accelerating at 1.2 ms-2 north along a straight road whilst
towing a 350 kg trailer. Friction forces on the car and trailer are 750 N and 250 N
respectively. Find:(a) the driving force acting on the drive wheels of the car?
(b)
The driving force will be the force causing the acceleration plus the force required to overcome
the resistances which is
F = ma = 1250 x 1.2 = 1500N plus (750N + 250N) = 2500N in total. This is because the
resistive forces must be overcome even BEFORE any acceleration can take place. This force must
also come from the car’s engine.
the tension in the towbar attached to the trailer
The towbar provides the means by which the trailer is pulled forward. Unless the trailer is
accelerating towards the car, this will equal the force that the trailer is exerting on it and,
consequently, on the car. This force must overcome the friction presented by the trailer and
cause the trailer to accelerate and therefore = ma = 350 x 1.2 plus the force required to
overcome the friction, so plus 250N. The total will therefore be 420 + 250 = 670 N
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5. If a 250 g metal block is dropped from 1.2 m onto a spring loaded buffer [spring
stiffness = 860 Nm-1], find:(a) the gravitational potential energy it has lost in falling 1.2 m
(b)
(c)
(d)
(e)
(f)
This equals mgh = 0.25 x 9.8 x 1.2 =
2.94 joules of energy
the kinetic energy of the 250 g mass just as it hits the spring
The gravitational potential energy of the mass due to being 1.2 m higher than the spring has
been converted into kinetic energy which now equals 2.94 joules
the speed of the mass just as it hits the spring
2.94 joules = ½ mv2 = ½ x 0.25 x v2
so v2 = 2.94 x 2/0.25 = 23.52 so v = √23.52 = 4.85 msec-1 = approx. 4.9 msec-1
Alternatively, use v2 = u2 + 2ax so v2 = 0 + 2 x 9.8 x 1.2 = 23.52 and v = √23.52 = 4.9 msec-1
the elastic strain energy gained by the spring when the mass is stopped
This will equal the kinetic energy of the block just before hitting the spring which in turn will
equal the original potential energy of the block = 2.94 J
how far the spring is compressed in stopping the mass
Energy of compression = ½ kx2 so 2.94 = ½ x 860 x x2
so x2 = 2.94 x 2/860 = 0.0068 and x = √0.0068 = 0.083 m = 8.3 cm
----------------------------------------------------------------------------------6. A 48 g golf ball travelling south at 34 ms-1 hits a brick wall and rebounds north at
28 ms-1. If it is in contact with the wall for 0.098 seconds, find:(a) the ball’s change in velocity
(b)
Choose north as positive, south as negative. Change in velocity therefore equals 48 – (-34) =
82 msec-1 north
the impulse given to the wall by the ball
This is a conservation of momentum problem. The change in momentum of the ball equals the
impulse it experiences and therefore the impulse the impulse the ball gives to the wall.
Initial momentum = 0.048 x 34 = 1.632 kgmsec-1 south
Because we choose north as positive, south as negative,
The momentum after the collision is 0.048 x 28 = 1.344 kgmsec-1 north.
The change in momentum = 1.344 – (-1.632) = 2.976 kgmsec-1 north
(Or simply multiply the change in velocity (82) by the ball’s mass)
(c)
the net average force applied to the ball by the wall
F ∆ t = m ∆v
and so F (average) = 2.976/0.098 = 30.37 or approx. 30N (2 sig figs)
------------------------------------------------------------------------7. If a 125 g air track glider, fitted with a Velcro bumper, is travelling to the right at
1.45 ms-1 and collides with and attaches to a stationary 150 g air track glider, find:(a) the momentum of the 125 g glider before impact
Momentum = 0.125 x 1.45 = 0.181 kgmsec-1
(b)
the velocity of the gliders after the impact
Momentum must be conserved so total momentum afterwards = 0.181 kgmsec-1
Therefore 0.181 = (0.125 + 0.150) x v
So v = 0.181/0.275 = 0.66 msec-1
----------------------------------------------------------------------------------8. If a 1.12 tonne V8 supercar accelerates from rest to reach a velocity of 168 kmh-1 in
the first 5.2 seconds of a race, find:(a) the work done by the V8 engine to reach 168 kmh-1 in 5.2 seconds
The work done = the gain in kinetic energy.
Kinetic energy = ½ mv2 = ½ x 1120 x 168/3.6 = 26133 J or approx. 26.1 kJ
(b)
the power developed by the supercar in the first 5.2 seconds of the race
Power = rate of doing work = 261333/5.2 = 5026 Watts or 5.05 kW