A fireworks display is made from three masses (mA = mB = 100g, mc = 200g) as sho

Question

A fireworks display is made from three masses (mA = mB = 100g, mc = 200g) as shown. Gravity points down. (a) The fireworks display is held together and launched vertically upwards from the ground with initial velocity VA = VB = vc = 100 ms^-1. Determine the maximum height relative to ground of the three masses if joined for the whole duration (6 marks). (b) Now, the fireworks display is launched as described in part (a), but the masses then split. Just before the split, the velocity is given by VA = VB = Vc = 2ms^-1 j. Just after the split, the velocity of A is given by vA = -5ms^-1 i + 5ms^-1 j and the velocity of C is given by vc = -2 5ms^-1 j. Determine the velocity of B just after the split (8 marks). (c) Determine the maximum height relative to ground of the centre of mass after the split described in part (b) (2 marks). (d) Determine the total energy required for the split described in part (b) to occur (6 marks).

Explanation / Answer

ma = 0.1 kg

mb = 0.1 kg

mc  = 0.2 kg

a) v = 100 m/s

h = v^2 / 2g = 100^2 / (2*9.81) = 509.68 m

Maximum height = 509.68 m

b) momentum conservation

ma ua + mb ub + mc uc  = ma va + mb vb + mc vc

0.1 * 2 j + 0.1 * 2 j + 0.2 * 2 j = 0.1 * ( - 5 i + 5 j ) + 0.1 * vb  + 0.2*( - 2 j )

0.8 j = - 0.5 i + 0.5 j - 0.4 j + 0.1 vb

0.5 i + 0.7 j = 0.1 vb

vb = 5 i + 7 j

Velocity of B after split = 5 ms-1  i + 7 ms-1 j

c) Since momentum is conserved centre of mass doesn't change after split

maximum height of centre of mass = 509.68 m

d) Energy required for split to occur = Change in kinetic energy = 0.1*(5^2 + 5^2) + 0.1*(5^2 + 7^2) + 0.2*2^2

- 0.1*2^2 - 0.1*2^2 - 0.2*2^2 = 11.6 J

Energy required = 11.6 J

  

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.