Two balls X and Y are supported by long strings, as shown in Fig. 3.1. The balls are each pulled back and pushed towards each other.

Question 22

Two balls X and Y are supported by long strings, as shown in Fig. 3.1.

Fig. 3.1

The balls are each pulled back and pushed towards each other. When the balls collide at the position shown in Fig. 3.1, the strings are vertical. The balls rebound in opposite directions.

Fig. 3.2 shows data for X and Y during this collision.

Fig. 3.2

The positive direction is horizontal and to the right.

(a) Use the conservation of linear momentum to determine the mass M of Y. [3]

(b) State and explain whether the collision is elastic. [1]

(c) Use Newton’s second and third laws to explain why the magnitude of the change in momentum of each ball is the same. [3]

Reference: Past Exam Paper – June 2015 Paper 21 Q3

Solution:

(a)

{Momentum before collision = momentum after collision}

(4.5 × 50) + (– 2.8 × M) = (–1.8 × 50) + (1.4 × M)

(M = ) 75 g

(b) The total initial kinetic energy/KE is not equal to the total final kinetic energy/KE, so not elastic

(c) From Newton’s 3^rd law, the force on X is equal and opposite to the force on Y.

From Newton’s 2^nd law, the force equals/is proportional to rate of change of momentum (F = Δp/Δt)

Since the time of collision is the same for both balls, the change in momentum is also the same.

{Newton’s 2^nd law is about the ‘rate of change of momentum’ while the questions ask about the ‘change in momentum’. So, we need to make reference to the time of collision being the same to be able to compare the magnitude of change in momentum.}

State the type of field, or fields, that may cause a force to be exerted on a particle that is

Question 9

(a) State the type of field, or fields, that may cause a force to be exerted on a particle that is

(i) uncharged and moving, [1]

(ii) charged and stationary, [1]

(iii) charged and moving at right-angles to the field. [2]

(b) A particle X has mass 3.32 × 10^-26 kg and charge +1.60 × 10^-19 C.

The particle is travelling in a vacuum with speed 7.60 × 10⁴ m s^-1. It enters a region of uniform magnetic field that is normal to the direction of travel of the particle. The particle travels in a semicircle of diameter 12.2 cm, as shown in Fig. 6.1.

Fig. 6.1

For the uniform magnetic field,

(i) state its direction, [1]

(ii) calculate the magnetic flux density. [3]

(c) A second particle Y has mass less than that of particle X in (b) and the same charge.

It enters the region of uniform magnetic field in (b) with the same speed and along the same initial path as particle X.

On Fig. 6.1, draw the path of particle Y in the region of the magnetic field. [1]

Reference: Past Exam Paper – June 2015 Paper 41 & 43 Q6

Solution:

(a)

(i) gravitational

{Note that a particle will have mass and so, a gravitational force would act.}

(ii) gravitational and electric

(iii) magnetic, gravitational and electric

{A magnetic force is exerted only when the particle is moving.}

(b)

(i) out of (plane of) paper/page

{Using Fleming’s right hand rule.}

(ii)

{Since the particle moves in a circle, there is a centripetal force.

The centripetal force is provided by the magnetic force.

Magnetic force = Centripetal force

Bqv = mv² / r}

B = mv / qr                                                                             

B = (3.32×10^-26 × 7.6×10⁴) / (1.6×10^-19 × 6.1×10^-2)  

B = 0.26 T                                                                              

(c)

sketch: semicircle with diameter < 12.2 cm

{B = mv / qr

Radius r = mv / Bq

When the mass m is smaller, the radius of the circular path is also smaller.}