Question 23
A stationary firework explodes into three pieces. The masses and the velocities of the three pieces immediately after the explosion are shown.
What are speed v1 and speed v2?
Reference: Past Exam Paper – March 2017 Paper 12 Q10
Solution:
Answer: B.
From the conservation of momentum,
Sum of momentum before explosion = Sum of momentum after explosion
The momentum before the explosion is zero as the firework was stationary.
Momentum is a vector quantity, so we need to consider the directions.
The momentum vector can be broken down into 2 components: horizontal and vertical.
Consider the vertical components:
{Momentum = mv}
Downward momentum = 100 × 8 = 800 g ms-1
Sum of upward momentum = 50v1sin 60° + 50v2 sin 60°
Sum of upward momentum = downward momentum
50v1 sin 60° + 50v2 sin 60° = 800
(v1 + v2) 50sin 60° = 800
v1 + v2= 800 / 50sin 60°
v1 + v2= 18.48 eqn (1)
Consider the horizontal components,
50 × v1 cos 60° = 50 × v2 cos 60°
v1 = v2
So, the speeds v1and v2 are equal. Consider equation (1) again,
v1 + v1= 18.48 (since v1= v2)
2v1 = 18.48
Speed v1 = 18.48 / 2 = 9.24 m s-1
No comments:
Post a Comment