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QuestionQ4 (10 marks) PH10 10∘10^{\circ} (a) A 5.00−g5.00-\mathrm{g} bullet moving with an initial speed of 400 m/s400 \mathrm{~m} / \mathrm{s} is fired into and passes through a 1.00−kg1.00-\mathrm{kg} block as shown in Figure 4(a). The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 900 N/m900 \mathrm{~N} / \mathrm{m}. The block moves a maximum distance of 5.00 cm to the right after the bullet exits the block.
Figure 4(a) (i) Find the speed at which the bullet emerges from the block.
ANS: (ii) Find the mechanical energy converted into internal energy in the collision.
ANS: Note: Question 4 continues on page 9

Studdy Solution

STEP 1

1. The bullet and block interaction is an inelastic collision.
2. The spring follows Hooke's Law.
3. The surface is frictionless.
4. The system is isolated, conserving momentum during the collision.

STEP 2

1. Use conservation of momentum to find the speed of the bullet as it exits the block.
2. Use energy conservation to find the mechanical energy converted into internal energy.

STEP 3

Apply the conservation of momentum principle. Before the collision, the momentum is only due to the bullet. After the collision, both the bullet and block have momentum:
mbvbi=mbvbf+mBvB m_b v_{b_i} = m_b v_{b_f} + m_B v_B
Where: - mb=5.00 g=0.005 kg m_b = 5.00 \, \text{g} = 0.005 \, \text{kg} is the mass of the bullet. - vbi=400 m/s v_{b_i} = 400 \, \text{m/s} is the initial speed of the bullet. - vbf v_{b_f} is the final speed of the bullet. - mB=1.00 kg m_B = 1.00 \, \text{kg} is the mass of the block. - vB v_B is the speed of the block after the bullet exits.
STEP_1.1: Solve for vB v_B using the maximum compression of the spring:
12kx2=12mBvB2 \frac{1}{2} k x^2 = \frac{1}{2} m_B v_B^2
Where: - k=900 N/m k = 900 \, \text{N/m} is the spring constant. - x=0.05 m x = 0.05 \, \text{m} is the maximum compression of the spring.
vB=kx2mB v_B = \sqrt{\frac{k x^2}{m_B}}
vB=900×0.0521.00 v_B = \sqrt{\frac{900 \times 0.05^2}{1.00}}
vB=2.25 v_B = \sqrt{2.25}
vB=1.5 m/s v_B = 1.5 \, \text{m/s}
STEP_1.2: Substitute vB v_B back into the momentum equation to solve for vbf v_{b_f} :
0.005×400=0.005×vbf+1.00×1.5 0.005 \times 400 = 0.005 \times v_{b_f} + 1.00 \times 1.5
2=0.005vbf+1.5 2 = 0.005 v_{b_f} + 1.5
0.5=0.005vbf 0.5 = 0.005 v_{b_f}
vbf=0.50.005 v_{b_f} = \frac{0.5}{0.005}
vbf=100 m/s v_{b_f} = 100 \, \text{m/s}

STEP 4

Calculate the mechanical energy converted into internal energy. The initial kinetic energy minus the final kinetic energy gives the energy converted:
Initial kinetic energy:
KEinitial=12mbvbi2 KE_{\text{initial}} = \frac{1}{2} m_b v_{b_i}^2
=12×0.005×4002 = \frac{1}{2} \times 0.005 \times 400^2
=400 J = 400 \, \text{J}
Final kinetic energy:
KEfinal=12mbvbf2+12mBvB2 KE_{\text{final}} = \frac{1}{2} m_b v_{b_f}^2 + \frac{1}{2} m_B v_B^2
=12×0.005×1002+12×1.00×1.52 = \frac{1}{2} \times 0.005 \times 100^2 + \frac{1}{2} \times 1.00 \times 1.5^2
=25+1.125 = 25 + 1.125
=26.125 J = 26.125 \, \text{J}
Mechanical energy converted into internal energy:
ΔE=KEinitial−KEfinal \Delta E = KE_{\text{initial}} - KE_{\text{final}}
=400−26.125 = 400 - 26.125
=373.875 J = 373.875 \, \text{J}
The speed at which the bullet emerges from the block is 100 m/s \boxed{100 \, \text{m/s}} .
The mechanical energy converted into internal energy is 373.875 J \boxed{373.875 \, \text{J}} .

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