Saturday, December 30, 2017

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

Thursday, December 28, 2017

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

Monday, December 25, 2017

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
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