Question

A 12.0 g wad of sticky clay is hurled horizontally at a 100 g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After the impact, the block slides 7.5 m before coming to rest. If the coefficient of friction between the block and the surface is 0.650, what was the speed of the clay immediately before the impact

Answer 1

The speed of the clay immediately before the impact is 91.23 m/s

Step-by-step 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)

`U_1 = ((m_1 + m_2)V)/(m_1)` ------- 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

`(K_f -K_i) + (F_k*d) = 0\\\\-K_i +F_k*d = 0\\\\K_i = F_k*d \\\\(1)/(2) (m_1+m_2)v_i^2 = \mu_k (m_1 +m_2)gd\\\\(1)/(2)v_i^2 = \mu_kgd\\\\v_i^2 = 2 \mu_kgd\\\\v_i = √(2 \mu_kgd)`

`v_i` is the common velocity of the clay-block system immediately after the impact, which is equal to V in equation (i)

`U_1 = ((m_1+m_2)V)/(m_1) = ((m_1+m_2)v_i)/(m_1)\\\\U_1 = (m_1+m_2)/(m_1)(√(2 \mu_kgd))\\\\U_1 = (0.012+0.1)/(0.012)(√(2 *0.65*9.8*7.5))\\\\U_1 = 9.3333(9.77497)\\\\U_1 = 91.23 \ m/s`

Therefore, the speed of the clay immediately before the impact is 91.23 m/s