Sunday, December 9, 2018

A rocket of mass 30 000 kg sits on a launch pad on the Earth’s surface. The rocket motors provide an upward force of 330 kN on the rocket.


Question 8
A rocket of mass 30 000 kg sits on a launch pad on the Earth’s surface. The rocket motors provide an upward force of 330 kN on the rocket.

What is the initial acceleration of the rocket?
A 0.12 m s-2                      B 1.1 m s-2                        C 1.2 m s-2                        D 11 m s-2





Reference: Past Exam Paper – November 2015 Paper 13 Q11





Solution:
Answer: C.

The forces acting on the rocket are its weight and the upward force provided by the rocket motors.

Weight of rocket = mg = 30 000 × 9.81
Weight = 294 300 N


The initial acceleration is provided by the resultant of the two forces.

Resultant force = ma

Upward force – Weight = ma
330 000 – 294 300 = 30 000 a

Acceleration a = 35 700 / 30 000 = 1.19 m s-2 ≈ 1.2 m s-2

Saturday, December 8, 2018

A stationary firework explodes into three different fragments that move in a horizontal plane, as illustrated in Fig. 2.1.


Question 10
(a) State the principle of conservation of momentum. [2]


(b) A stationary firework explodes into three different fragments that move in a horizontal plane, as illustrated in Fig. 2.1.


Fig. 2.1

The fragment of mass 3.0M has a velocity of 7.0 m s-1 perpendicular to line AB.
The fragment of mass 2.0M has a velocity of 6.0 m s-1 at angle θ to line AB.
The fragment of mass 1.5M has a velocity of 8.0 m s-1 at angle θ to line AB.

(i) Use the principle of conservation of momentum to determine θ. [3]

(ii) Calculate the ratio
kinetic energy of fragment of mass 2.0M
kinetic energy of fragment of mass 1.5M
 [2]
 [Total: 7]





Reference: Past Exam Paper – June 2018 Paper 22 Q2





Solution:
(a) The principle of conservation of momentum states that the total momentum (of a system of bodies) is constant when there is no (resultant) external force.


(b)
(i)
(Momentum p =) mv   or (3.0M × 7.0)            or (2.0M × 6.0)            or (1.5M × 8.0)           

{The firework was stationary before the explosion. So, the momentum before collision is zero. In other words, the momentum should cancel out.}

{Vertically, (we need to find the vertical components of the fragments where necessary)
Sum of momentum upwards = Sum of momentum downwards}

3.0M × 7.0 = (2.0M × 6.0 sinθ) + (1.5M × 8.0 sinθ)   
{21 = 12 sinθ + 12 sinθ
24 sinθ = 21
θ = sin-1 (21/24)}

θ = 61°           


(ii)
(Kinetic energy E =) ½ mv2                    
ratio = (½ × 2.0M × 6.02) / (½ × 1.5M × 8.02)
ratio = 0.75                                         
Currently Viewing: Physics Reference |