Answer:
The speed of the clay immediately before the impact is 91.23 m/s
Explanation:
Given;
mass of clay, m₁ = 12g = 0.012 kg
mass of wooden block, m₂ = 100g = 0.1 kg
initial velocity of the wooden block, u₂ = 0
distance moved by the wooden block, d = 7.5 m
coefficient of friction, μk = 0.65
Apply the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = v(m₁ +m₂)
where;
u₁ is the initial velocity of the clay immediately before the impact
v is the common velocity of clay-block system after impact
u₂ = 0
m₁u₁ = v(m₁ +m₂)
[tex]U_1 = \frac{(m_1 + m_2)V}{m_1}[/tex] ------- equ. (i)
Apply the principle of conservation of energy after the impact
ΔK + ΔU = 0
where;
ΔK is change in kinetic energy
ΔU is change in internal energy of the system due to frictional force
[tex](K_f -K_i) + (F_k*d) = 0\\\\-K_i +F_k*d = 0\\\\K_i = F_k*d \\\\\frac{1}{2} (m_1+m_2)v_i^2 = \mu_k (m_1 +m_2)gd\\\\\frac{1}{2}v_i^2 = \mu_kgd\\\\v_i^2 = 2 \mu_kgd\\\\v_i = \sqrt{2 \mu_kgd}[/tex]
[tex]v_i[/tex] is the common velocity of the clay-block system immediately after the impact, which is equal to V in equation (i)
[tex]U_1 = \frac{(m_1+m_2)V}{m_1} = \frac{(m_1+m_2)v_i}{m_1}\\\\U_1 = \frac{m_1+m_2}{m_1}(\sqrt{2 \mu_kgd})\\\\U_1 = \frac{0.012+0.1}{0.012}(\sqrt{2 *0.65*9.8*7.5})\\\\U_1 = 9.3333(9.77497)\\\\U_1 = 91.23 \ m/s[/tex]
Therefore, the speed of the clay immediately before the impact is 91.23 m/s
