QuestionQ4 (10 marks) PH10 $10^{\circ}$ (a) A $5.00-\mathrm{g}$ bullet moving with an initial speed of $400 \mathrm{~m} / \mathrm{s}$ is fired into and passes through a $1.00-\mathrm{kg}$ 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 $900 \mathrm{~N} / \mathrm{m}$. 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
Calculate the mechanical energy converted into internal energy. The initial kinetic energy minus the final kinetic energy gives the energy converted:
Initial kinetic energy:
$$KE_{\text{initial}} = \frac{1}{2} m_b v_{b_i}^2$$
$$= \frac{1}{2} \times 0.005 \times 400^2$$
$$= 400 \, \text{J}$$
Final kinetic energy:
$$KE_{\text{final}} = \frac{1}{2} m_b v_{b_f}^2 + \frac{1}{2} m_B v_B^2$$
$$= \frac{1}{2} \times 0.005 \times 100^2 + \frac{1}{2} \times 1.00 \times 1.5^2$$
$$= 25 + 1.125$$
$$= 26.125 \, \text{J}$$
Mechanical energy converted into internal energy:
$$\Delta E = KE_{\text{initial}} - KE_{\text{final}}$$
$$= 400 - 26.125$$
$$= 373.875 \, \text{J}$$
The speed at which the bullet emerges from the block is $\boxed{100 \, \text{m/s}}$.
The mechanical energy converted into internal energy is $\boxed{373.875 \, \text{J}}$.