Download 15th September

15th September.
Questions obtained from Fundamentals of Physics Extended, Halliday, Resnick and Walker.
TEST (A2) - Energy and Momentum.
1.
A 2kg sloth steps off a tree and drops 5m to the ground.
a)
What is the sloth’s gravitational potential energy relative to the ground?
b)
Just before landing all the sloth’s gravitational potential energy is converted to
kinetic energy. What is the speed of the sloth, just before landing?
2.
The spring of a gun is compressed a distance d of 3.2cm from its relaxed state, and a
ball of mass m=12g is put in the barrel. With what speed will the ball leave the barrel
once the gun is fired? The spring constant is 7.5N/cm. Assume no friction and a
horizontal gun barrel. The spring’s potential energy is given by U  12 kd 2 .
3.
A child of mass m is released from rest at the top of a three-dimensional water slide at
a height h=8.5m above the level of the pool. How fast is the child moving when she
reaches the pool? Assume that the slide is frictionless because of the water on it.
4.
A 61kg bungee jumper is on a bridge 45m above a river. In its relaxed state, the
elastic bungee cord has length L=25m. Assume that the cord obeys Hooke’s law
( F  kx , where k is the elastic constant and x the displacement from its relaxed state),
with k=160N/m.
5.
a)
If the jumper stops before reaching the water, what is the height h of her feet
above the water at her lowest point?
b)
What is the net force on her at her lowest point?
A cart with mass 340g moving on a frictionless linear track an initial speed of 1.2m/s
strikes a second cart of unknown mass at rest. The collision between the carts is
elastic. After the collision, the first cart continues in its original direction at 0.66m/s.
a)
What is the mass of the second cart?
b)
What is its speed after impact?
6.
A machine gun fires 0.05kg bullets at a speed of 1000m/s. The gunner, holding the
machine gun in his hands, can exert an average force of 180N against the gun.
Determine the maximum number of bullets he can fire per minute while still holding
the gun steady.
7.
Two titanium spheres approach each other head-on with the same speed and collide
elastically. After the collision, one of the spheres, whose mass is 300g, remains at
rest. What is the mass of the other sphere?
TEST (A2) - Energy and Momentum.
1.
A 2kg sloth steps off a tree and drops 5m to the ground.
a)
What is the sloth’s gravitational potential energy relative to the ground?
b)
Just before landing all the sloth’s gravitational potential energy is converted to
kinetic energy. What is the speed of the sloth, just before landing?
ANSWER:
1.
a)
Just a straightforward use of E p  mgh , so that E p  2  9.8  5  98J
b)
They tell you that all of the GPE is converted to KE so:
Ek  E p
 12 mv 2  mgh
v

2 gh
98
 9.9ms 1
2.
The spring of a gun is compressed a distance d of 3.2cm from its relaxed state, and a
ball of mass m=12g is put in the barrel. With what speed will the ball leave the barrel
once the gun is fired? The spring constant is 7.5N/cm. Assume no friction and a
horizontal gun barrel. The spring’s potential energy is given by U  12 kd 2 .
ANSWER:
2.
Again, just equate the potential energy with kinetic energy (care with the units!):
Ek  E p
 12 mv 2  12 kd 2
v d
k
m
 0.032
 8ms 1
750
0.012
3.
A child of mass m is released from rest at the top of a three-dimensional water slide at
a height h=8.5m above the level of the pool. How fast is the child moving when she
reaches the pool? Assume that the slide is frictionless because of the water on it.
ANSWER:
3.
It matters not that the slide is three-dimensional! Just equate GPE with KE (it’s a
conservative field), so:
E p  Ek
 mgh  12 mv 2
v

2 gh
2  9.8  8.5
 12.9ms 1
4.
A 61kg bungee jumper is on a bridge 45m above a river. In its relaxed state, the
elastic bungee cord has length L=25m. Assume that the cord obeys Hooke’s law
( F  kx , where k is the elastic constant and x the displacement from its relaxed state),
with k=160N/m.
a)
If the jumper stops before reaching the water, what is the height h of her feet
above the water at her lowest point?
b)
What is the net force on her at her lowest point?
ANSWER:
4.
a)
She stops so there is no KE, some of the GPE she had must be stored in the
bungee cord as elastic potential energy, but she will still have some GPE.
45mg  12 kd 2  mg 45  L  d 
 0  12 kd 2  mg L  d 
 0  12 kd 2  mgd  mgL
We now have a quadratic in d, which we solve using the formula (ignoring the
negative value for d):
0  12 kd 2  mgd  mgL
d 
61  9.8 
61  9.82  2  160  61  9.8  25
160
 17.9m
So the height of her feet above the water is just 45  25  17.9  2.1m .
b)
The two forces acting on her are her weight and the force modelled by
Hooke’s Law. The net force is therefore just:
F  mg  kd
 61  9.8  160  17.9
 2270 N
i.e the force acts to lift the jumper back in the air.
5.
A cart with mass 340g moving on a frictionless linear track an initial speed of 1.2m/s
strikes a second cart of unknown mass at rest. The collision between the carts is
elastic. After the collision, the first cart continues in its original direction at 0.66m/s.
a)
What is the mass of the second cart?
b)
What is its speed after impact?
ANSWER:
5.
Its an elastic collision so both momentum and KE is conserved, so
mAu A  mB u B  mA v A  mB vB
and
1
2
m A u A2  12 mB u B2  12 m A v A2  12 mB v B2
 m A u A2  mB u B2  m A v A2  mB v B2
Substituting values into the momentum equation:
0.340  1.2  0.340  0.66  mB v B
 mB v B  0.1836
And into the KE equation:
0.340  1.2 2  0.340  0.66 2  mB v B2
mB v B2  0.341
Now its easier to answer part (b) first, since equating the two, we have:
0.1836v B  0.341
 v B  1.86ms 1
From which we get m B 
0.1836
 0.0987 , or just 99gms
1.86
6.
A machine gun fires 0.05kg bullets at a speed of 1000m/s. The gunner, holding the
machine gun in his hands, can exert an average force of 180N against the gun.
Determine the maximum number of bullets he can fire per minute while still holding
the gun steady.
ANSWER:
6.
This is just a question on impulse. Remember force is just rate of change of
momentum with respect to time, so we require: nmv  F where n represents the
number of bullets per second. Plugging in the numbers, and multiplying by 60 to get
rounds per minute:
n
180  60
 216
0.05  1000
7.
Two titanium spheres approach each other head-on with the same speed and collide
elastically. After the collision, one of the spheres, whose mass is 300g, remains at
rest. What is the mass of the other sphere?
ANSWER:
7.
Again an elastic collision, so both momentum and KE is conserved, so
mAu A  mB u B  mA v A  mB vB
and
1
2
m A u A2  12 mB u B2  12 m A v A2  12 mB v B2
 m A u A2  mB u B2  m A v A2  mB v B2
Substituting values into the momentum equation, letting A be the sphere at rest after
the collision:
0.3u  mB u  mB v B
 0.3u  mB v B  u 
And into the KE equation:
0.3u 2  mB u 2  mB v B2

 0.3u 2  mB v B2  u 2

Dividing the two equations:


0.3u 2 m B v B2  u 2

0.3u
m B v B  u 
u
v B  u v B  u 
v B  u 
 2u  v B
And substituting back:
0.3u  mB v B  u 
  0.3u  3umB
2u  v B

 mB  0.1kg