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

Answer 2

The speed of the beanbag when it strikes the crate is 34.46 m/s.

Inelastic collision:

In an inelastic collision, two bodies collide, stick together and move with the same velocity. The momentum is conserved while the kinetic energy is not.

As the beanbag sticks to the crate they move together with the same velocity until they stop after 0.935 m due to frictional force.

Let M be the combined mass of the crate and beanbag. the frictional force acting on the system is:

$F=-\mu_k Mg\\\\Ma=-\mu_k Mg\\\\a=-\mu g\\\\a=-0.480\times9.8\;m/s^2\\\\a=4.7\;m/s^2$will be the acceleration of the system.

Let the initial velocity of the system be u and the final velocity is 0, then from the third equation of motion:

$0=u^2-2as\\\\u=\sqrt[]{2as}\\\\u=\sqrt{2\times4.7\times0.935}\;m/s\\\\u=2.96\;m/s$

So the momentum after collision is

Mu = (0.354 + 3.77)×2.96 = 12.2 kgm/s

Which must be equal to the initial momentum that is contributed only by the beanbag, since the crate is on rest. Let the speed of the bean bag be v then initial momentum should be:

12.2 = 0.354 × v

v = 34.46 m/s

Learn more about inelastic collision:

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