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YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Sunday, October 18, 2020

A metal ball of mass 40 g falls vertically onto a spring, as shown in Fig. 4.1.

Question 17

A metal ball of mass 40 g falls vertically onto a spring, as shown in Fig. 4.1.

Fig. 4.1 (not to scale)

 

The spring is supported and stands vertically. The ball has a speed of 2.8 m s-1 as it makes contact with the spring. The ball is brought to rest as the spring is compressed.

 

(a) Show that the kinetic energy of the ball as it makes contact with the spring is 0.16 J. [2]

 

 

(b) The variation of the force F acting on the spring with the compression x of the spring is shown in Fig. 4.2.

Fig. 4.2

 

The ball produces a maximum compression XB when it comes to rest. The spring has a spring constant of 800 N m-1.

 

Use Fig. 4.2 to

(i) calculate the compression XB, [2]

 

(ii) show that not all the kinetic energy in (a) is converted into elastic potential energy in the spring. [2]

 

 

 

 

 

Reference: Past Exam Paper – June 2015 Paper 21 Q4

 

 

 

 

 

Solution:

(a)

kinetic energy = ½ mv2            

kinetic energy = ½ × 0.040 × (2.8)2 = 0.157 J or 0.16 J

 

 

(b)

(i)

{From Hooke’s law,}

k = F / x                       or F = kx        

 

{x = F / k}

XB = 14 / 800

XB = 0.0175 m                                    

 

(ii)

{The elastic potential energy stored in the spring can be obtained from the area under the F-x graph.}

 

area under graph = elastic potential energy stored

or Elastic PE = ½ kx2                 or ½ Fx

 

{Elastic PE = ½ Fx = ½ × 14 × 0.0175}

(energy stored =) 0.1225 J less than KE (of 0.16 J)

 

{This value of elastic PE is less than the KE calculated above.}

Sunday, October 11, 2020

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}

Sunday, October 4, 2020

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}

ΔEp = mgΔh

ΔEp = 0.602 × 9.8 × 0.086

ΔEp = 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, Ek = ½ mv2

Ek = ½ × 0.002 ×3902

Ek = 152 J        or 153 J           or 150 J

 

 

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

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