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
