A bullet of mass 2.0 g is fired horizontally into a block of wood of mass 600 g.

 Question 43

A bullet of mass 2.0 g is fired horizontally into a block of wood of mass 600 g. The block is suspended from strings so that it is free to move in a vertical plane.

 

The bullet buries itself in the block. The block and bullet rise together through a vertical distance of 8.6 cm, as shown in Fig. 3.1.

 

Fig. 3.1

 

(a) (i) Calculate the change in gravitational potential energy of the block and bullet. [2]

 

(ii) Show that the initial speed of the block and the bullet, after they began to move off together, was 1.3 m s^-1. [1]

 

 

(b) Using the information in (a)(ii) and the principle of conservation of momentum, determine the speed of the bullet before the impact with the block. [2]

 

 

(c) (i) Calculate the kinetic energy of the bullet just before impact. [2]

 

(ii) State and explain what can be deduced from your answers to (c)(i) and (a)(i) about the type of collision between the bullet and the block. [2]

 

 

 

 

 

Reference: Past Exam Paper – June 2015 Paper 2 Q3

 

 

 

 

 

Solution:

(a) (i)

{Total mass = mass of bullet + mass of block = 2 + 600 = 602 g = 0.602 kg}

ΔE_p = mgΔh

ΔE_p = 0.602 × 9.8 × 0.086

ΔE_p = 0.51 J

 

 

(ii)

{KE of bullet and block = Change in gravitational PE

½ mv² = 0.51 J}

v² = (2 × 0.51) / 0.602

Initial speed of block and bullet, v = 1.3 m s^-1 

 

 

(b)

{Sum of momentum before impact = Sum of momentum after impact

Before the impact, the bullet is moving while the block is at rest.

Momentum of bullet before impact = mv = 2.0 × v

Sum of momentum before impact = 2v + 0 = 2v

After the impact, the bullet and the block move as a single body with the same speed.

Momentum after impact = (600+2) × 1.3

Sum of momentum before impact = Sum of momentum after impact }

2v = 602 × 1.3             (allow 600)

Speed of bullet, v = 390 m s^-1

 

 

(c) (i)

Kinetic energy of bullet, E_k = ½ mv²

E_k = ½ × 0.002 ×390²

E_k = 152 J        or 153 J           or 150 J

 

 

(ii) The kinetic energy is not the same. So, the collision is inelastic.

The maximum useful output power of a car travelling on a horizontal road is P. The total resistive force acting on the car is kv2, where v is the speed of the car and k is a constant.

Question 42

The maximum useful output power of a car travelling on a horizontal road is P. The total resistive force acting on the car is kv², where v is the speed of the car and k is a constant.

 

Which equation is correct when the car is travelling at maximum speed?

A v³ = P / k                 B v² = P / k                 C v = (P/k)²                 D v = (P/k)³

 

Reference: Past Exam Paper – November 2019 Paper 13 Q17

 

Solution:

Answer: A.

Power = P

Resistive force F = kv²

For a body moving with constant speed,

Power P = Force × velocity

P = F × v

P = (kv²) × v

P = kv³

Make v the subject of formula,

v³ = P / k

Distinguish between gravitational potential energy and electric potential energy.

Question 41

(a) Distinguish between gravitational potential energy and electric potential energy. [2]

(b) A body of mass m moves vertically through a distance h near the Earth’s surface. Use the defining equation for work done to derive an expression for the gravitational potential energy change of the body. [2]

(c) Water flows down a stream from a reservoir and then causes a water wheel to rotate, as shown in Fig. 4.1.

Fig. 4.1

As the water falls through a vertical height of 120 m, gravitational potential energy is converted to different forms of energy, including kinetic energy of the water. At the water wheel, the kinetic energy of the water is only 10% of its gravitational potential energy at the reservoir.

(i) Show that the speed of the water as it reaches the wheel is 15 m s^-1. [2]

(ii) The rotating water wheel is used to produce 110 kW of electrical power. Calculate

the mass of water flowing per second through the wheel, assuming that the production of electric energy from the kinetic energy of the water is 25% efficient. [3]

Reference: Past Exam Paper – November 2011 Paper 21 Q4

Solution:

(a) Electric potential energy is the energy (stored) when a charge is moved due to work done in an electric field while gravitational potential energy is the energy (stored) when a mass is moved due to work done in gravitational field.

(b)

Work done = Force × distance moved (in the direction of force)

and Force = mg                                                                                  

(GPE =) mg × h                      OR mg × Δh                                       

(c)

(i)

{As the water falls, GPE is converted into KE of the water.

However, the process is only 10% efficient. That is, only 10% (= 0.1) of the GPE is converted into KE.

0.1 × Gravitational potential energy = Kinetic energy}

0.1 × mgh = ½ mv²                

0.1 × m × 9.81 × 120 = ½ mv²

{m can be eliminated.

Speed v = √(2×0.1×9.81×120) }

Speed of water, v = 15.3 m s^-1

(ii)

Power, P = ½ mv² / t                                                              

{Power = energy / time

The falling water has kinetic energy

Kinetic energy = ½ mv²

Power = (½ mv²) / t = ½ × (m/t) × v²  

where m/t is the mass of water flowing per second}

{Only 25% of the kinetic energy produces electric energy.

Electrical power = 0.25 × ½ × (m/t) × v² = 110×10³ W

0.25 × ½ × (m/t) × v² = 110×10³

(m/t) = 110×10³ / (0.25×0.5×15.3²)}

Mass of water flowing per second: m/t = (110×10³) / (0.25×0.5×15.3²)

m/t = 3760 kg s^-1