An object of mass m travelling with speed v has a head-on collision with another object of mass m travelling with speed v in the opposite direction.

Question 25

An object of mass m travelling with speed v has a head-on collision with another object of mass m travelling with speed v in the opposite direction. The two objects stick together after the collision.

 

What is the total loss of kinetic energy in the collision?

A 0                   B ½ mv²                         C mv²                                                 D 2mv²

 

 

 

Reference: Past Exam Paper – November 2015 Paper 12 Q13

 

 

 

Solution:

Answer: C.

Momentum is a vector quantity and we need to consider the direction.

Since the objects are moving in opposite direction, one of them will have a negative value.

Sum of momentum before collision = mv – mv = 0

 

Kinetic energy is a scalar quantity and so, the direction of motion does not matter.

Sum of KE before collision = ½ mv² + ½ mv² = mv²

 

Momentum is always conserved.

Sum of momentum after collision = Sum of momentum before collision = 0

Sum of momentum after collision = 0

2mv_f = 0

Giving the speed after collision, v_f = 0

That is, the objects do not move after the collision.

KE after collision = 0

 

Loss in KE = mv² – 0 = mv²

Which statement about electromagnetic radiation is correct?

 Question 26

Which statement about electromagnetic radiation is correct?

A Waves of wavelength 5 × 10^-9 m are high-energy gamma rays.

B Waves of wavelength 3 × 10^-8 m are ultra-violet waves.

C Waves of wavelength 5 × 10^-7 m are infra-red waves.

D Waves of wavelength 9 × 10^-7 m are light waves.

 

 

 

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q22

 

 

 

Solution:

Answer: B.

Estimates for the different components of the electromagnetic spectrum should be known at A-Level.

 

All EM waves travel at the same speed of 3.0×10⁸ m s^-1.

Gamma rays have the highest frequencies, followed by X-rays and UV.

 

v = fλ

 

For the given wavelength of UV waves,

f = v / λ = 3.0×10⁸ / 3.0×10^-8 = 1×10¹⁶ Hz

 

This is a correct estimate for the frequency of UV waves.

A monochromatic plane wave of speed c and wavelength λ is diffracted at a small aperture.

 Question 36

A monochromatic plane wave of speed c and wavelength λ is diffracted at a small aperture.

 

The diagram illustrates successive wavefronts.

After what time will some portion of the wavefront GH reach point P?

A 3λ / 2c                      B 2λ / c                                    C 3λ / c                                    D 4λ / c

 

 

 

Reference: Past Exam Paper – March 2017 Paper 12 Q28

 

 

 

Solution:

Answer: C.

Speed of wave = c

The distance between two successive wavefronts is equal to a wavelength.

So, the distance between wavefront GH and the wavefront containing point P is equal to 3 wavelengths (= 3λ).

Speed = distance / time

Time = distance / speed

Time = 3λ / c