Question

A 12 kg block is at rest on a level floor. A 417 g glob of putty is thrown at the block such that it travels horizontally, hits the block, and sticks to it. The block and putty slide 15 cm along the floor. If the coefficient of sliding friction is 0.40, what is the initial speed of the putty?

Answer 1

Answer:

u₂ = 32.29 m/s

Explanation:

m₁ = 12 kg

u₁ = 0 m/s

m₂ = 417 g = 0.417 kg

d = 15 cm = 0.15 m

μ = 0.40

u₂ = ?

It is an inelastic collision. After the collision

We can apply ∑ F = m*a  for the system (m₁ + m₂)

where   m = m₁ + m₂ = 12 kg + 0.417 kg = 12.417 kg

then

∑ F = - Ff = m*a

if

Ff = μ*N = μ*W = μ*(m*g) = μ*m*g

⇒    - μ*m*g = m*a    ⇒    a = - μ*g = - 0.40*9.8 m/s² = - 3.92 m/s²

⇒    a = - 3.92 m/s²

Since vf = 0 m/s (for the system)

we use the equation

vf² = vi²+2*a*d   ⇒   0 = vi²+2*a*d   ⇒    vi = √(-2*a*d)

⇒    vi = √(-2*(- 3.92 m/s²)*0.15 m)

⇒    vi = 1.0844 m/s

we can use the equation for an inelastic collision

m₁*u₁ + m₂*u₂ = (m₁ + m₂)*vs

since m = m₁ + m₂;  u₁ = 0 m/s   and   vs = vi

we have

m₁*0 + m₂*u₂ = m*vi

⇒   u₂ = m*vi / m₂

⇒   u₂ = 12.417 kg*1.0844 m/s / 0.417 kg

⇒   u₂ = 32.29 m/s   (→)