Question

For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italian ham. The slices of ham are weighed on a plate of mass 0.400 kg placed atop a vertical spring of negligible mass and force constant of 200 N/m . The slices of ham are dropped on the plate all at the same time from a height of 0.250 m . They make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (SHM). You may assume that the collision time is extremely small.
a. What is the amplitude of oscillation A of the scale after the slices of ham land on the plate?
b. What is the period of oscillation T of the scale?

Answer 1

Answer with Explanation:

We are given that

$m_1=0.3 kg$

$m_2=0.4 kg$

Spring constant,$k=200 N/m$

$h=0.250 m$

a.Speed of ham before collision

$u=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.25}=2.2m/s$

Collision is inelastic

According to law of conservation of momentum

$m_1u_1=(m_1+m_2)v$

$0.3\times 2.2=(0.3+0.4)v$

$v=\frac{0.3\times 2.2}{0.7}=0.94 m/s$

Kinetic energy of masses=Spring potential energy at maximum distance

$\frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}kA^2$

$A^2=\frac{(m_1+m_2)v^2}{k}$

$A=\sqrt{\frac{(m_1+m_2)v^2}{k}}$

Substitute the values

$A=\sqrt{\frac{(0.3+0.4)\times (0.94)^2}{200}}=0.056 m=0.056\times 100=5.6 cm$

1 m=100 cm

b.Time period,$T=2\pi\sqrt{\frac{(m_1+m_2)}{k}}$

$T=2\pi\sqrt{\frac{(0.3+0.4)}{200}}=0.37 s$