Electromagnetic waves of wavelength λ and frequency f travel at speed c in a vacuum. What describes the wavelength and speed of electromagnetic waves of frequency f / 2?

Question 3

Electromagnetic waves of wavelength λ and frequency f travel at speed c in a vacuum.

What describes the wavelength and speed of electromagnetic waves of frequency f / 2?

Wavelength                 Speed in a vacuum

A          λ / 2                                         c / 2

B          λ / 2                                         c

C         2λ                                            c

D         2λ                                            2c

Reference: Past Exam Paper – June 2013 Paper 13 Q24

Solution:

Answer: C.

The speed of EM waves in vacuum is c.

Speed c = f λ

Wavelength λ = c / f

The speed of all EM waves in vacuum is constant (= c).

If the frequency is now f/2

Speed in vacuum = c

New wavelength = c / (f/2) = 2c / f

 

Since λ = c / f,

New wavelength = 2 λ

A lead pellet of mass 10.0 g is shot horizontally into a stationary wooden block of mass 100 g. The pellet hits the block with an impact velocity of 250 m s-1.

Question 3

A lead pellet of mass 10.0 g is shot horizontally into a stationary wooden block of mass 100 g. The pellet hits the block with an impact velocity of 250 m s^-1. It embeds itself in the block and it does not emerge.

What will be the speed of the block immediately after the pellet is embedded?

A 23 m s^-1                          B 25 m s^-1                          C 75 m s^-1                         D 79 m s^-1

Reference: Past Exam Paper – November 2013 Paper 13 Q13

Solution:

Answer: A.

From the conservation of momentum,

Sum of momentum before collision = Sum of momentum of collision

Momentum = mass × velocity = mv

Sum of momentum before collision = (10×10^-3 × 250) + (100×10^-3 × 0)

Sum of momentum before collision = (10×10^-3 × 250)

After collision, the pellet is embedded into the block. Both can now be considered as a single body of (100+10 =) 110 g. Let the speed with which they move = v.

Momentum after collision = (100+10) ×10^-3 × v

Sum of momentum before collision = Sum of momentum of collision

(10×10^-3 × 250) = (100+10) ×10^-3 × v

Speed v = (10 × 250) / 110 = 22.7 = 23 m s^-1

A truck of mass 500 kg moves from rest at the top of a section of track 400 m long and 30 m high, as shown. The frictional force acting on the truck is 250 N throughout its journey.

Question 6

A truck of mass 500 kg moves from rest at the top of a section of track 400 m long and 30 m high, as shown. The frictional force acting on the truck is 250 N throughout its journey.

What is the final speed of the truck?

A 14 m s^-1                         B 24 m s^-1                         C 31 m s^-1                         D 190 m s^-1

Reference: Past Exam Paper – November 2016 Paper 11 & 13 Q19

Solution:

Answer: A.

From the conservation of energy,

GPE of truck at top = KE of truck at bottom + Work done against friction

GPE of truck at top = mgh = 500 × 9.81 × 30 = 147 150 J

Work done against friction = F × d = 250 × 400 = 100 000 J

GPE of truck at top = KE of truck at bottom + Work done against friction

147 150 = KE of truck at bottom + 100 000

KE of truck at bottom = 147 150 – 100 000 = 47 150 J

½ mv² = 47 150

Final speed v = √(2×47150 / 500) = 13.7 m s^-1 ≈ 14 m s^-1