MOMENTUM
5 marks · HardA fireworks shell of mass 0.75 kg is initially at rest. It explodes in mid-air into two fragments: fragment X (0.45 kg) flies west at 16 m/s and fragment Y (unknown mass) flies east.
Calculate the mass of fragment Y produced in the explosion.
Calculate the velocity of fragment Y after the explosion.
Explain why fragment Y moves in the opposite direction to fragment X.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) 0.30 kg · must show massY = 0.75 - 0.45
(b) 24 m/s (east) m/s · must show momentum before = momentum after = 0, correct sign for opposite directions, p = m × v
(c) Because total momentum must remain zero, so the two fragments move in opposite directions with equal but opposite momentum. · must show momentum conserved, fragments have equal and opposite momenta
Common mistakes: Wrong mass for Y; Ignoring direction of velocities
Show mark scheme and worked solution
Marking points
A1 (a) · 1 mark · Correct subtraction to find mass of Y: 0.75 - 0.45 = 0.30 kg.
M1 (b) · 1 mark · Applies conservation of momentum and sets up the correct equation with direction. (follow-through)
M2 (b) · 1 mark · Solves equation for velocity of Y using correct mass and momentum direction. (follow-through)
A2 (b) · 1 mark · Final answer for velocity of Y: 24 m/s east (allow sign or direction). (follow-through)
B1 (c) · 1 mark · Mentions momentum is conserved and fragments must move with equal and opposite momentum.
Worked solution
(a) Mass of Y:
(b) Take west as negative, east as positive. System starts at rest so total momentum is 0.
Fragment Y moves east at 24 m/s.
(c) The shell was at rest, so by conservation of momentum, the two fragments must have equal and opposite momenta. This means they move in opposite directions.