# Chapter 10 - Energy - Exercises and Problems - Page 273: 32

(a) The speed of the brick after the collision is $0.0476~v_0$ (b) 95.2% of the mechanical energy was lost in the collision.

#### Work Step by Step

Let $m_c$ be the mass of the clay. Let $m_b$ be the mass of the brick. (a) We can use conservation of momentum to set up an equation for this perfectly inelastic collision. $p_f=p_0$ $(m_c+m_b)~v_f = m_c~v_0$ $v_f = \frac{m_c~v_0}{m_c+m_b}$ $v_f = \frac{(0.050~kg)~v_0}{0.050~kg+1.0~kg}$ $v_f = 0.0476~v_0$ The speed of the brick after the collision is $0.0476~v_0$ (b) We can find the initial kinetic energy. $KE_0 = \frac{1}{2}m_cv_0^2$ $KE_0 = \frac{1}{2}(0.050~kg)v_0^2$ $KE_0 = 0.025~v_0^2$ We can find the final kinetic energy. $KE_f = \frac{1}{2}(m_c+m_b)(0.0476~v_0)^2$ $KE_f = \frac{1}{2}(0.050~kg+1.0~kg)(0.0476~v_0)^2$ $KE_f = 0.0012~v_0^2$ We can find the percentage of kinetic energy that remains after the collision. $\frac{KE_f}{KE_0}\times 100\% = \frac{0.0012~v_0^2}{0.025~v_0^2}\times 100\% = 4.8\%$ We can find the percentage of kinetic energy that was lost in the collision. $100\% - 4.8\% = 95.2\%$ 95.2% of the mechanical energy was lost in the collision.

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