Question

A 0.460 kg block of wood rests on a horizontal frictionless surface and is attached to a spring (also horizontal) with a 22.5 N/m force constant that is at its equilibrium length. A 0.0600 kg wad of Play-Doh is thrown horizontally at the block with a speed of 2.90 m/s and sticks to it. Determine the amount by which the Play-Doh-block system compresses the spring.

Answer 1

Answer:

0.05086 m

Explanation:

$m_1$ = Mass of block = 0.46 kg

$m_2$ = Mass of Play-doh = 2.9 kg

$u_1$ = Initial Velocity of block = 0 m/s

$u_2$ = Initial Velocity of Play-doh = 0.06 m/s

v = Velocity of combined mass

k = Spring constant = 22.5 N/m

x = Compression of spring

As the linear momentum of the system is conserved

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow v=\frac{0.46\times 0 + 0.06\times 2.9}{0.46 + 0.06}\\\Rightarrow v=0.33461\ m/s$

The kinetic energy of motion and the spring will balance each other

$\frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}kx^2\\\Rightarrow (m_1+m_2)v^2=kx^2\\\Rightarrow x=\sqrt{\frac{(m_1+m_2)v^2}{k}}\\\Rightarrow x=\sqrt{\frac{(0.46+0.06)\times 0.33461^2}{22.5}}\\\Rightarrow x=0.05086\ m$

The amount by which the spring gets compresses is 0.05086 m