TEST - Energy and Momentum

TEST out of 40 marks

[one mark] which of the following express the Joule in its fundamental units?

A) kg m² s^-2

B) kg m s^-2

C) kg m s^-1

D) N s

The correct answer for one mark is A

[one mark] A ball is thrown vertically upwards. Resistance is negligible. What is the variation with time t of the gravitational potential energy E of the ball?

For one mark the correct answer is C

[one mark] A boat with an output engine power of 15,000 W moves through water out of speed of 20 m s^-1. What is the resistance force acting on the boat?

For one mark the correct answer is 0.75 kN

[one mark] a motor of input power 160 W raises a mass of 16.0 kg vertically at a constant speed of 0.50 m s^-1. What is the efficiency of the system?

The correct answer for one mark is 50%

[one mark] A system that consists of a single spring stores a total elastic potential energy E when a load is added to the spring. Another identical spring connected in a parallel is added to the system. The same load is now applied to the parallel springs. What is the elastic potential energy store in one of the parallel springs in the changed system?

For one mark the correct answer is C

[one mark] A graph shows the variation of force acting on an object moving in a straight line with distance moved by the object. Which area represents the work done on the object during its motion from O to Q.

The correct answer for one mark is D

[one mark] An object is released from rest at X and slides to Y. The vertical distance between X and Y is 10 m. During the motion, 20% of the objects initial gravitational potential energy is lost as friction. What is the speed of the object at Y?

The correct answer for one mark is C

[one mark] An object of mass 1.0 kg hangs at rest from the spring. The spring has a negligible mass and the spring constant K is 20 N m ^-1. What is the elastic potential energy stored in the spring?

For one mark the correct answer is B

[one mark] A projectile is launched upwards on an angle theta to the horizontal with an initial momentum, P and an initial energy E. Resistance is negligible what are the momentum and total energy of the projectile at the highest point in the motion?

For one mark the correct answer is D

[one mark] Two objects, M1 and M2 approach each other along a straight line with speeds V1 and V2 as shown. The objects collide and stick together what is the total change of linear momentum of the objects as a result of the collision?

The correct answer for one mark is A

[one mark] A tennis ball is dropped from rest from a height. It hits the ground and bounces back to a lower height. Resistance is negligible. What is correct about the collision of the tennis ball with the ground?

A) elastic because momentum of the system is conserved

B) elastic because the kinetic energy of the system is conserved

C) inelastic because momentum of the system is not conserved

D) in elastic because the kinetic energy of the system is not conserved

For one mark the correct answer is D

[one mark] Elastic motor is used to lift a heavy load. The sankey diagram shows the energy transformations involved in the process. What is the efficiency of the entire process?

The correct answer for one mark is B

Question one - the graph shows the variation with time T of the horizontal force F exerted on a tennis ball by a racket. The tennis ball was moving at 4.5 m s^-1 before contact with the racket. The mass of the tennis ball is 5.8×10^-2 kg. The area under the curve 0.94 N s. [three marks] calculate the speed of the ball as it leaves the racket.

To calculate the speed of the ball as it leaves the racket, we can use the impulse-momentum theorem, which states that the change in momentum is equal to the impulse applied to it.

1. Calculate the initial momentum of the tennis ball:

   Initial momentum = mass × initial velocity= (5.8 × 10^-2 kg) × (4.5 m/s)= 0.261 kg m/s

2. Calculate the impulse applied (area under the force-time graph):Impulse = 0.94 N s

3. The final momentum will be:Final momentum = Initial momentum + Impulse= 0.261 kg m/s + 0.94 N s= 0.261 kg m/s + 0.94 kg m/s= 1.201 kg m/s

4. Calculate the final velocity:Final momentum = mass × final velocity1.201 kg m/s = (5.8 × 10^-2 kg) × final velocityFinal velocity = 1.201 kg m/s / (5.8 × 10^-2 kg)= 20.69 m/s

Thus, the speed of the ball as it leaves the racket is approximately 20.69 m/s. And rounded to be 21 ms - don't forget direction for vector quantities

Question one continued - [4 marks] determine with reference to the work done by the average force the horizontal distance travelled by the ball while it was in contact with the racket.

Question one continued - [two marks] explain, in terms of impulse, why follow through is so important in sports.

Follow through is vital in sports because it allows an athlete to apply force over a longer time during impact. This increased time leads to a greater impulse (force x time), resulting in a larger change in momentum. As a result, the object can achieve higher speed or distance.

Question to a small metal pendulum bob of mass 75 g is suspended at rest from a fixed point with a length of thread of negligible mass. Air resistance is negligible. The bob is then displaced to the left. At time T equals zero the bob is moving horizontally to the right at 0.80 m s ^-1. It collides with a small stationary object also of mass 110 g both objects that move together with motion that is simple harmonic.

[two marks] state the law of conservation of momentum

The law of conservation of momentum states that in a closed system, the total momentum before an event (such as a collision) is equal to the total momentum after the event. In formula terms, this can be expressed as: Total momentum before collision = Total momentum after collision.

Question 2 continued - [three marks] calculate the velocity of the combine masses immediately after the collision.

To calculate the velocity of the combined masses after the collision, we use the law of conservation of momentum. The initial momentum before the collision can be calculated as:

Initial momentum = (mass of the pendulum bob) × (velocity of the pendulum bob) + (mass of the stationary object) × (velocity of the stationary object) = (0.075 kg) × (0.80 m/s) + (0.110 kg) × (0) = 0.060 kg m/s

After the collision, they stick together, so the total mass is: Combined mass = 0.075 kg + 0.110 kg = 0.185 kg

Let V be the final velocity after the collision: Total momentum after collision = combined mass × final velocity Using conservation of momentum: 0.060 kg m/s = (0.185 kg) × V

Solving for V gives: V = 0.060 kg m/s / 0.185 kg ≈ 0.324 m/s

Thus, the velocity of the combined masses immediately after the collision is approximately 0.324 m/s.

Question two continued - [three mark] with reference to kinetic energy, show that the collision is inelastic

In an inelastic collision, the total kinetic energy before the collision is greater than the total kinetic energy after the collision. Before the collision, the kinetic energy of the pendulum bob is: KE_initial = 0.5 × mass × velocity² = 0.5 × 0.075 kg × (0.80 m/s)² = 0.024 J. The stationary object has no kinetic energy, so KE_initial_total = 0.024 J. After the collision, the combined mass moves together with a velocity of approximately 0.324 m/s. The kinetic energy after the collision is: KE_final = 0.5 × (0.185 kg) × (0.324 m/s)² ≈ 0.00987 J. Since KE_initial_total (0.024 J) > KE_final (0.00987 J), the collision is inelastic because kinetic energy is not conserved.

Question three - an elastic climbing rope is tested by fixing one end of the rope to the top of a crane. The other end of the rope is connected to a block, which is initially position a. The block is released from rest. The mass of the rope is negligible. The unexpected length of the rope is 65.0 m from position a to position be the block falls freely. - [one mark] State the law of conservation of energy.

The law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In a closed system, the total energy remains constant.

Question three continued - [two marks] at position B the rope starts to extend. Determine the speed of the block at position, be using the law of conservation of energy.

Given:

• Block falls a height of 65.0 m.

• Starts from rest, so initial speed is 0.

Step-by-Step:

1. Conservation of Energy: At position A, the block has only gravitational potential energy:

   EA=mgh

   At position B, the block has kinetic energy and the rope has elastic potential energy:

   Eb=12mv^2+12kΔx^2

   Since total energy is conserved, EA=EB

2. Simplify if Rope Stretch is Small: If the rope doesn't stretch much, assume elastic potential energy is negligible:

   mgh=12mv^2

   Solve for v:

   v=2gh

3. Substitute Values:

   v= Square root of 2×9.81 m/s^2×65.0 m = 35.7 m/s

So, the speed of the block at position B is approximately 35.7 m/s.

Question three continued - [three marks] calculate the magnitude of the average force exerted by the rope on the block between B and C. Include a free body diagram for the block in your solution.

1. Work-Energy Theorem:

   ΔKE=1/2m(v²C−v²B)

   Assuming vC=0 m/svC​=0m/s (the block nearly stops at C), and vB=35.7 m/sv:

   ΔKE=1/2(5.8×10^−2 kg)×(0−35.7²) ΔKE=−36.9J

2. Average Force: Using d=1 md=1m for the rope stretch:

   Favg=ΔKE/d=−36.9 J/1 m=−36.9 N

   Magnitude: Favg=36.9 N

Free Body Diagram:

• Forces on the Block:

  • Gravitational force Fg=mgFg​=mg (downward).

  • Tension force FropeFrope​ (upward, from the stretched rope).

The rope exerts an upward force, opposing the block's motion.

Question three continued - [two marks] the length reached by the rope at sea is 77.4 m. Using energy, consider considerations, determine the elastic constant of the rope.

Given:

• Speed at position B: vB=35.7 m/svB​=35.7m/s

• Speed at position C: vC=0 m/svC​=0m/s

• Rope stretches by ΔL=12.4 mΔL=12.4m (from 65.0 m to 77.4 m).

Step-by-Step:1. Change in Kinetic Energy:ΔKE=1/2m(v²C−v²B) ΔKE=12(5.8×10^−2 kg)×(0−35.7²)=−36.9 J

Elastic Potential Energy:

The work done by the rope is stored as elastic potential energy:

Elastic=1/2kΔL²

Equating change in kinetic energy to elastic potential energy:

1/2k(12.4)²=36.9

k×76.88=36.9

Okay=36.9/76.88 =0.48 N/m

Conclusion:

The elastic constant kk of the rope is 0.48 N/m.