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 = 50v1
sin 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 v1
and 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
mass should be converted from grams to kg
ReplyDeletethere is no need in this case as we are not calculating the actual momentum. rather, we are equating terms.
Deleteeven if we had converted the mass, it would cancel on both sides