An electron is travelling in a vacuum at a speed of 3.4 × 107 m s-1. The electron enters a region of uniform magnetic field of flux density 3.2 mT, as illustrated in Fig. 8.1.

 Question 17

An electron is travelling in a vacuum at a speed of 3.4 × 10⁷ m s^-1. The electron enters a region of uniform magnetic field of flux density 3.2 mT, as illustrated in Fig. 8.1.

Fig. 8.1

 

The initial direction of the electron is at an angle of 30° to the direction of the magnetic field.

 

(a) When the electron enters the magnetic field, the component of its velocity vN normal to the direction of the magnetic field causes the electron to begin to follow a circular path.

 

Calculate:

(i) vN [1]

(ii) the radius of this circular path. [3]

 

 

(b) State the magnitude of the force, if any, on the electron in the magnetic field due to the component of its velocity along the direction of the field. [1]

 

 

(c) Use information from (a) and (b) to describe the resultant path of the electron in the magnetic field. [1]

[Total: 6]

 

 

Reference: Past Exam Paper – June 2019 Paper 42 Q8

 

Solution:

(a)

(i)

vN = 3.4×10⁷ × sin 30°

vN = 1.7 × 10⁷ m s^-1

 

 

(ii)

{The magnetic force provides the centripetal force.

Magnetic force = Bqv}

 

mv² / r = Bqv               or r = mv / Bq

 

{Making r the subject of formula,

r = mv / Bq }

 

r = (9.11×10^-31 × 1.7×10⁷) / (3.2×10^-3 × 1.60×10^-19)

r = 0.030 m                                                                

 

 

(b) zero

{It is only the component of the velocity which is perpendicular to the field that causes a force.}

 

 

(c) helix/coil

 

{The motion can be described in two components:

- circular motion in one plane (due to the component perpendicular to the field)

- linear motion in the third direction}

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.

Wires are used to connect a battery of negligible internal resistance to a lamp, as shown in Fig. 7.1.

Question 19

(a) Define the ohm. [1]

 

 

(b) Wires are used to connect a battery of negligible internal resistance to a lamp, as shown in Fig. 7.1.

Fig. 7.1

 

The lamp is at its normal operating temperature. Some data for the filament wire of the lamp and for the connecting wires of the circuit are shown in Fig. 7.2.

 

Fig. 7.2

 

(i) Show that

resistance of filament wire     

total resistance of connecting wires = 1000.

[2]

 

 

(ii) Use the information in (i) to explain qualitatively why the power dissipated in the filament wire of the lamp is greater than the total power dissipated in the connecting wires. [1]

 

 

(iii) The lamp is rated as 12 V, 6.0 W. Use the information in (i) to determine the total resistance of the connecting wires. [3]

 

 

(iv) The diameter of the connecting wires is decreased. The total length of the connecting wires and the resistivity of the metal of the connecting wires remain the same.

 

State and explain the change, if any, that occurs to the resistance of the filament wire of the lamp. [3]

[Total: 10]

 

 

 

 

 

Reference: Past Exam Paper – November 2017 Paper 21 Q7

 

 

 

 

 

Solution:

(a) (the ohm is defined as) volt / ampere

 

 

(b)

(i)

{Resistance of wire:}

R = ρL / A

 

{Cross-sectional area A = πd² / 4

 

Resistance of filament wire = ρL / (πd² /4)

Resistance of connecting wires = 0.028ρ × 7.0L / {π (14d)² / 4}

}

 

ratio = [ρL / (πd² /4)] / [0.028ρ × 7.0L / {π (14d)² / 4}] = 1000

 

or

ratio = 14² / (0.028 × 7) = 1000

 

 

(ii)

The same current flows (in the connecting and filament wires) but the lamp/filament (wire) has greater resistance

 

{Power dissipated = I²R}

 

 

(iii)

P = V² / R                    or P = VI                      or P = I²R

 

{Make R the subject formula in whichever you choose.

 

R = V² / P                    or P = I² / R                 or R = V / I  and I = P / V = 6/12 = 0.5 A}

 

(for filament wire) R = 12² / / 6.0        or R = 6.0 / 0.50²        or R = 12 / 0.50

(for filament wire) R = 24 Ω

 

{From the ratio of (b)(i), the ratio of

Resistance of filament wire / total resistance of connecting wires = 1000

 

Total resistance of connecting wires = resistance of filament wire / 1000

Total resistance of connecting wires = 24 / 1000}

 

(for connecting wire) R = 24 / 1000 = 2.4 × 10^–2 Ω

 

 

(iv)

{Resistance of wire: R = ρL / A

When the diameter is decreased, the cross-sectional area A decreases.}

 

The resistance of the connecting wire increases.

 

{The total resistance in the circuit increases.

From Ohm’s law, I = V / R

When the resistance increases, the current in the circuit decreases.}

 

The current in circuit/lamp/filament (wire) decreases.

or

potential difference across lamp/filament (wire) decreases

 

{Now, considering the FILAMENT lamp.

V = IR

Since the current has decreased, the p.d. across the lamp decreases.}

 

(So) the resistance of lamp/filament (wire) decreases.

 

{When the p.d. across a filament lamp increases, its resistance also increases. In the same way, when the p.d. across the filament lamp decreases, its resistance also decreases.}