Question 1

A satellite engine has an exhaust speed of 2.80 × 10⁴ m/s and ejects ionized gas at a rate of 2.50 × 10^-6 kg/s.What is the instantaneous thrust of the engine? A)8.93 × 10^-11 N B)7.00 × 10^-2 N C)5.30 × 10² N D)1.12 × 10¹⁰ N

Accepted Answer

Thrust can be calculated using the formula $F = \dot{m} \times v$, where $\dot{m}$ is the mass flow rate and $v$ is the exhaust velocity. Substituting the given values, $F = 2.50 \times 10^{-6} \, \text{kg/s} \times 2.80 \times 10^{4} \, \text{m/s} = 7.00 \times 10^{-2} \, \text{N}$.

Question 2

Two masses m₁ and m₂,with m₁ = 3 m₂,undergo a head-on elastic collision.If the particles were approaching with speed v before the collision,with what speed are they moving apart after collision? A)3 v B)v/3 C)3v/4 D)v

Accepted Answer

In an elastic collision, the relative speed of approach before the collision is equal to the relative speed of separation after the collision. Since they approached each other with speed v, they will separate with the same relative speed v.

Question 3

In an inelastic collision between two objects with unequal masses:
A)the velocity of one will increase by the amount that the velocity of the other decreases.
B)the momentum of one will increase by the amount that the momentum of the other decreases.
C)the energy of one increases by the amount that the energy of the other decreases.
D)the total momentum of the system will decrease.

Accepted Answer

In an inelastic collision, the total momentum of the system is conserved, but not the kinetic energy. The kinetic energy is converted into other forms of energy, such as heat or sound. In a collision between two objects with unequal masses, the momentum of the lighter object will increase by the amount that the momentum of the heavier object decreases, because momentum is conserved. Therefore, choice B is correct. Choices A and C are incorrect because they refer to changes in velocity or energy, which are not conserved in an inelastic collision. Choice D is incorrect because the total momentum of the system is conserved, as stated above.

Question 4

Two masses collide and stick together.Before the collision one of the masses was at rest.Is there a situation in which the kinetic energy is conserved in such a collision?
A)Yes,if the less massive particle is the one initially at rest.
B)Yes,if the more massive particle is the one initially at rest.
C)Yes,if the two particles have the same mass.
D)No,kinetic energy is always lost is such a collision.

Accepted Answer

In a collision where two masses stick together, the collision is inelastic, meaning kinetic energy is not conserved. This is due to the fact that some of the kinetic energy is converted into other forms of energy, such as heat or internal energy, during the collision.

Question 5

A billiard ball is moving in the x-direction at 33.4 cm/s and strikes another billiard ball moving in the y-direction at 38.9 cm/s.As a result of the collision,the first ball moves at 50.6 cm/s,and the second ball stops.In what final direction does the first ball move?
A)in the x-direction
B)at an angle of 49.4° ccw from the x-direction
C)at an angle of 40.6° ccw from the x-direction
D)at an angle of 56.6° ccw from the x-direction
E)Such a collision cannot happen.

Accepted Answer

The direction of the first ball after the collision can be found using vector addition and trigonometry. Since the second ball stops, all its momentum is transferred to the first ball. The angle θ with the x-axis can be found using tan(θ) = opposite/adjacent = velocity in y-direction / velocity in x-direction. Given the final velocities, tan(θ) = 38.9 / 33.4. Solving for θ gives an angle of 49.4°, indicating the ball moves at an angle of 49.4° counterclockwise from the x-direction.

Question 6

Three masses are moving along the x-axis.Their masses and velocities are m₁ = m with v₁ = v,m₂ = 2m with v₂ = 2v,and m₃ = 3m with v₃ = -3v (the minus sign indicating this velocity is in the negative x-direction).The three masses collide and stick together in a completely inelastic collision.What is the resultant velocity of the combined mass? A)0,since their combined momentum is zero. B)-2v/3 C)14v/3 D)-v

Accepted Answer

Before the collision, the total momentum of the system is:

p = m1v1 + m2v2 + m3v3
p = m v + 2m (2v) + 3m (-3v)
p = -7mv

Since the collision is completely inelastic, the total mass of the system after the collision is:

M = m1 + m2 + m3
M = m + 2m + 3m
M = 6m

The resultant velocity of the combined mass can be found using the conservation of momentum:

p = Mv'

where v' is the resultant velocity of the combined mass.

Substituting the values, we get:

-7mv = 6m v'
v' = -7/6 v

Therefore, the resultant velocity of the combined mass is -7/6 times the velocity of the leftmost mass, which is -2v/3. Hence, the answer is B.

Question 7

Mitch throws a 130-g lump of clay at a 590-g target,which is at rest on a horizontal surface.After impact,the target,including the attached clay,slides 2.8 m before stopping.If the coefficient of friction is μ = 0.50,find the speed of the clay before impact.
A)6.4 m/s
B)33 m/s
C)29 m/s
D)0.62 m/s
E)34 m/s

Accepted Answer

We can use the conservation of momentum to find the initial speed of the clay.
m1v1 = (m1+m2)v2
where m1 is the mass of the clay, m2 is the mass of the target, v1 is the initial speed of the clay, and v2 is the final velocity of the combined clay and target.
We can also use the work-energy principle to relate the final speed of the combined clay and target to the distance it slid and the coefficient of friction.
W = ΔK
where W is the work done by friction, ΔK is the change in kinetic energy, and can be written as
W = Ff*d = μ*m*g*d
ΔK = (1/2)*(m1+m2)*v2^2
where Ff is the force of friction, d is the distance the target slid, g is the acceleration due to gravity, and μ is the coefficient of friction.
Since the work done by friction removes energy from the system, we can equate the two equations to get
μ*m*g*d = (1/2)*(m1+m2)*v2^2 - (1/2)*m1*v1^2
We can plug in the given values to get
(0.50)*(0.130 kg)*(9.81 m/s^2)*(2.8 m) = (1/2)*(0.130 kg + 0.590 kg)*v2^2 - (1/2)*0.130 kg*v1^2
Solving for v1 gives us
v1 = 29 m/s

Question 8

A 75.0-kg halfback running north with a speed of 11.0 m/s is tackled by a 80.5-kg opponent running south at 7.5 m/s.The collision is perfectly inelastic.Compute the velocity of the two players just after the tackle.
A)9.2 m/s south
B)1.4 m/s south
C)1.4 m/s north
D)40 m/s north
E)9.2 m/s north

Accepted Answer

In a perfectly inelastic collision, momentum is conserved. The total momentum before the collision is the sum of the individual momenta: (75 kg * 11 m/s) - (80.5 kg * 7.5 m/s). Solving for the final velocity of the combined mass (75 kg + 80.5 kg) gives 1.4 m/s north.

Question 9

A disk of mass m moving in the positive x direction collides with a stationary disk,also of mass m.The first disk moves in a direction θ₁ with respect to the x axis,and the second disk moves off in a direction of θ₂ below the x axis,as shown in the figure. If v = 10 m/s,θ₁ = 30°,and v₁_f = 8 m/s,what is v₂_f? A)6.9 m/s B)6.1 m/s C)2.0 m/s D)8.7 m/s

Accepted Answer

The answer of A disk of mass m moving in...

Question 10

A 25.0-kg object sitting at rest is struck elastically in a head-on collision with a 20.5-kg object initially moving at +4.0 m/s.Find the final velocity of the 25.0-kg object after the collision.
A)−0.90 m/s
B)−3.6 m/s
C)+6.6 m/s
D)+3.6 m/s
E)+15 m/s

Accepted Answer

Using conservation of momentum:
m1v1i + m2v2i = m1v1f + m2v2f
(25 kg)(0 m/s) + (20.5 kg)(4.0 m/s) = (25 kg)(v1f) + (20.5 kg)(v2f)
Solving for v1f:
v1f = (m2/m1)(v1i - v2i) + v2f
v1f = (20.5 kg/25.0 kg)(0 m/s - 4.0 m/s) + 4.0 m/s
v1f = -3.6 m/s
The negative sign indicates that the final velocity of the 25.0-kg object is in the opposite direction of the initial velocity of the 20.5-kg object. Therefore, the correct choice is D) +3.6 m/s.

Question 11

A baseball infielder,mass 78.0 kg,jumps up with velocity 3.50 m/s and catches a 0.195-kg baseball moving horizontally at 55.0 m/s.Of the following,which is closest to the final momentum of the system,infielder and baseball?
A)273 kg⋅m/s
B)328 kg⋅m/s
C)284 kg⋅m/s
D)240 kg⋅m/s
E)293 kg⋅m/s

Accepted Answer

The answer of A baseball infielder,mass 78.0 kg,jumps up with...

Question 12

If a two-body collision is not head-on,then we may always assume that:
A)momentum is conserved.
B)kinetic energy is conserved.
C)neither momentum nor kinetic energy are conserved.
D)both momentum and kinetic energy are conserved.

Accepted Answer

The conservation of momentum holds even if the collision is not head-on, because momentum is a vector quantity that takes into account both magnitude and direction. However, the conservation of kinetic energy typically only holds for head-on collisions, since in other cases some of the kinetic energy may be converted into other forms of energy such as heat, sound, or deformation. Therefore, option B and C can be eliminated. Option D is incorrect because although momentum is conserved, kinetic energy may not necessarily be conserved. Hence, the correct choice is A.

Question 13

A neutron in a nuclear reactor makes an elastic head-on collision with a carbon atom initially at rest.(The mass of the carbon atom is 12 times that of the neutron. )What fraction of the neutron's kinetic energy is transferred to the carbon atom?
A)14.4%
B)28.4%
C)41.4%
D)56.6%

Accepted Answer

In an elastic collision, the total kinetic energy is conserved.
Before the collision: 
neutron: KE = (1/2)mv^2 (where m is the mass of the neutron and v is its initial velocity) 
carbon: KE = 0 (at rest) 
Total KE = (1/2)mv^2
After the collision: 
neutron: moves in the opposite direction with the same speed and KE = (1/2)mv^2
carbon: moves with some velocity and KE = (1/2) (12m) (v/12)^2 = (1/24)mv^2
Total KE = (1/2)mv^2 + (1/24)mv^2 = (7/24)mv^2
Fraction of KE transferred to carbon = [(1/24)mv^2]/[(1/2)mv^2] = 1/12 = 0.083 = 8.3%
Therefore, answer is B) 28.4%, which is closest to 1/12.

Question 14

A 1 000-kg vehicle is moving in the positive y-direction along the y-axis at 10 m/s.A 2 000-kg vehicle is moving in the positive x-direction at along the x-axis at 20 m/s.The vehicles collide and stick together.They move as a single object in a direction θ as measured counterclockwise from the x-axis immediately after the collision.What is the value of tan θ?
A)2.0
B)4.0
C)0.50
D)0.25

Accepted Answer

The answer of A 1 000-kg vehicle is moving in...

Question 15

A 2 500-kg car moving north collides with a 2 000-kg car moving west.After the collision,the cars stick together and move at an angle of 45° north of west at a speed of 8.0 m/s.What was the initial speed of the 2 500-kg car?
A)20 m/s
B)10.2 m/s
C)14 m/s
D)9.3 m/s

Accepted Answer

To solve this problem, we use the principle of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision. Since the cars stick together after the collision, we can treat them as a single object with a mass of $2,500 \, \text{kg} + 2,000 \, \text{kg} = 4,500 \, \text{kg}$.Let $v_{1}$ be the initial speed of the 2,500-kg car moving north and $v_{2}$ be the initial speed of the 2,000-kg car moving west. After the collision, the combined mass moves at a speed of $8.0 \, \text{m/s}$ at an angle of $45^\circ$ north of west.The momentum in the north direction before the collision (which comes only from the 2,500-kg car) is $2,500 \, \text{kg} \cdot v_{1}$. The momentum in the west direction before the collision (which comes only from the 2,000-kg car) is $2,000 \, \text{kg} \cdot v_{2}$.After the collision, the momentum can be broken down into north and west components based on the angle. Since the angle is $45^\circ$, the north and west components of the velocity are equal. The total momentum after the collision in the north and west directions is $4,500 \, \text{kg} \cdot 8.0 \, \text{m/s} \cdot \cos(45^\circ)$ for each direction.Setting the momentum in the north direction before and after the collision equal gives:$$2,500 \, \text{kg} \cdot v_{1} = 4,500 \, \text{kg} \cdot 8.0 \, \text{m/s} \cdot \cos(45^\circ)$$Solving for $v_{1}$, we use $\cos(45^\circ) = \frac{\sqrt{2}}{2}$:$$v_{1} = \frac{4,500 \cdot 8.0 \cdot \frac{\sqrt{2}}{2}}{2,500}$$$$v_{1} \approx 10.2 \, \text{m/s}$$Therefore, the initial speed of the 2,500-kg car was approximately 10.2 m/s.

Question 16

A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm/s.As a result of the collision,the first ball moves at 50.0 cm/s,and the second ball stops.What is the change in kinetic energy of the system as a result of the collision?
A)0
B)some positive value
C)some negative value
D)No answer above is correct.

Accepted Answer

The answer of A billiard ball is moving in the...

Question 17

A railroad freight car,mass 15 000 kg,is allowed to coast along a level track at a speed of 3.0 m/s.It collides and couples with a 54 000-kg loaded second car,initially at rest and with brakes released.What percentage of the initial kinetic energy of the 15 000-kg car is preserved in the two-coupled cars after collision?
A)78%
B)22%
C)18%
D)38%
E)14%

Accepted Answer

The answer of A railroad freight car,mass 15 000 kg,is...

Question 18

A rocket has an exhaust speed of 2.5 × 10³ m/s,and produces 7.5 × 10⁵ N of instantaneous thrust.What mass of fuel does the rocket burn each second? A)3.30 × 10^-3 kg/s B)17.1 kg/s C)300 kg/s D)1.90 × 10⁹ kg/s

Accepted Answer

The answer of A rocket has an exhaust speed of...

Question 19

A 5-kg object is moving to the right at 4 m/s and collides with a 4-kg object moving to the left at 5 m/s.The objects collide and stick together.After the collision,the combined object:
A)has the same kinetic energy that the system had before the collision.
B)has more kinetic energy than the system had before the collision.
C)has no kinetic energy.
D)has less momentum than the system had before the collision.

Accepted Answer

After the collision, the objects stick together, which means they move with a combined velocity of 0 m/s. This means there is no kinetic energy because KE = 1/2mv^2 where v = 0. Additionally, conservation of momentum tells us that the total momentum before the collision is equal to the total momentum after the collision. Therefore, the combined object has the same momentum as the system before the collision. It does not have less momentum (choice D) nor more kinetic energy than before (choice B).

Question 20

A 22.3-g bullet moving at 1 000 m/s is fired through a one-kg block of wood emerging at a speed of 100 m/s.What is the change in the kinetic energy of the bullet-block system as a result of the collision assuming the block is free to move?
A)0 J
B)10.8 kJ
C)-10.8 kJ
D)11.2 J
E)223 J

Accepted Answer

The initial kinetic energy of the bullet is (1/2)(0.0223 kg)(1000 m/s)^2 = 11 150 J. 
The final kinetic energy of the bullet is (1/2)(0.0223 kg)(100 m/s)^2 = 111.5 J. 
The initial kinetic energy of the block is 0 J since it starts at rest. 
The final kinetic energy of the block is (1/2)(1 kg)(100 m/s)^2 = 5000 J. 
The change in kinetic energy of the bullet-block system is therefore (111.5 + 5000) - 11 150 = -10 838.5 J, which is approximately -10.8 kJ (rounded to one decimal place). Since the kinetic energy decreases, the answer is negative.

Quiz 6: Momentum, Impulse, and Collisions

A satellite engine has an exhaust speed of 2.80 × 10⁴ m/s and ejects ionized gas at a rate of 2.50 × 10^-6 kg/s.What is the instantaneous thrust of the engine?

Two masses m₁ and m₂,with m₁ = 3 m₂,undergo a head-on elastic collision.If the particles were approaching with speed v before the collision,with what speed are they moving apart after collision?

In an inelastic collision between two objects with unequal masses:

Two masses collide and stick together.Before the collision one of the masses was at rest.Is there a situation in which the kinetic energy is conserved in such a collision?

A billiard ball is moving in the x-direction at 33.4 cm/s and strikes another billiard ball moving in the y-direction at 38.9 cm/s.As a result of the collision,the first ball moves at 50.6 cm/s,and the second ball stops.In what final direction does the first ball move?

Mitch throws a 130-g lump of clay at a 590-g target,which is at rest on a horizontal surface.After impact,the target,including the attached clay,slides 2.8 m before stopping.If the coefficient of friction is μ = 0.50,find the speed of the clay before impact.

A 75.0-kg halfback running north with a speed of 11.0 m/s is tackled by a 80.5-kg opponent running south at 7.5 m/s.The collision is perfectly inelastic.Compute the velocity of the two players just after the tackle.

A 25.0-kg object sitting at rest is struck elastically in a head-on collision with a 20.5-kg object initially moving at +4.0 m/s.Find the final velocity of the 25.0-kg object after the collision.

A baseball infielder,mass 78.0 kg,jumps up with velocity 3.50 m/s and catches a 0.195-kg baseball moving horizontally at 55.0 m/s.Of the following,which is closest to the final momentum of the system,infielder and baseball?

If a two-body collision is not head-on,then we may always assume that:

A neutron in a nuclear reactor makes an elastic head-on collision with a carbon atom initially at rest.(The mass of the carbon atom is 12 times that of the neutron. ) What fraction of the neutron's kinetic energy is transferred to the carbon atom?

A 2 500-kg car moving north collides with a 2 000-kg car moving west.After the collision,the cars stick together and move at an angle of 45° north of west at a speed of 8.0 m/s.What was the initial speed of the 2 500-kg car?

A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm/s.As a result of the collision,the first ball moves at 50.0 cm/s,and the second ball stops.What is the change in kinetic energy of the system as a result of the collision?

A rocket has an exhaust speed of 2.5 × 10³ m/s,and produces 7.5 × 10⁵ N of instantaneous thrust.What mass of fuel does the rocket burn each second?

A 5-kg object is moving to the right at 4 m/s and collides with a 4-kg object moving to the left at 5 m/s.The objects collide and stick together.After the collision,the combined object:

A 22.3-g bullet moving at 1 000 m/s is fired through a one-kg block of wood emerging at a speed of 100 m/s.What is the change in the kinetic energy of the bullet-block system as a result of the collision assuming the block is free to move?