Question

A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanbag is moving exactly horizontally when it strikes and sticks to the inside wall of the crate, which is lined with velcro. As a result of this impact, the crate slides backwards 0.935 meters along the concrete, which has a coefficient of kinetic friction of \mu_k = 0.480μ ​k ​​ =0.480. The beanbag has a mass of 0.354 kg and the empty crate has a mass of 3.77 kg. Calculate the speed of the beanbag vv at the moment it strikes the crate. You may neglect the effect of air resistance.

Answer 1

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

$m_1$ = Mass of bean bag = 0.354 kg

$m_2$ = Mass of empty crate = 3.77 kg

$v_1$ = Speed of the bean bag

$v_2$ = Speed of the crate

Acceleration

$a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g$

$a=--9.81\times 0.48=4.7088\ m/s^2$

From equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s$

In this system the momentum is conserved

$m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s$

The speed of the bean bag is 31.42383 m/s