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