# Collision question?

## A: Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s? B: How far does block two travel, d2 in meters, before coming to rest after the collision?

Apr 15, 2017

I get
A. $11.29 m {s}^{-} 1$
B. $24.05 m$
Results rounded to two decimal places.

#### Explanation:

A. Initial velocity of block of mass ${m}_{1} = 6.25 m {s}^{-} 1$
During sliding movement impeding force of friction
$F = {\mu}_{k} R$
where $R$ is normal reaction and is $= m g$
$\therefore$ Force of friction $F = {\mu}_{k} {m}_{1} g$
Deceleration produced due to this force$= \frac{F}{m} _ 1 = \frac{{\mu}_{k} {m}_{1} g}{m} _ 1 = {\mu}_{k} g$
$a = - 0.27 \times 9.81 = - 2.6487 m {s}^{-} 2$ .....(1)

using the kinematic expression
${v}^{2} - {u}^{2} = 2 a s$ ......(2)
${v}^{2} - {\left(6.5\right)}^{2} = 2 \left(- 2.6487\right) 2.4$
$v = 5.43 m {s}^{-} 1$
Momentum of mass ${m}_{1}$ just before collision$= {m}_{1} v = 74.37 k g m {s}^{-} 1$

Given that first block stops after its collision with second block. Due to Law of conservation of momentum, momentum of second block just after the collision $= 74.37 k g m {s}^{-} 1$
Velocity of second mass $= \frac{\text{momentum}}{m} _ 2 = \frac{74.37}{6.5} = 11.29 m {s}^{-} 1$

B. Using (1) for Deceleration due friction also for block ${m}_{2}$, and using kinematic expression (2) we get
${0}^{2} - {\left(11.29\right)}^{2} = 2 \left(- 2.6487\right) {d}_{2}$
$\implies {d}_{2} = {\left(11.29\right)}^{2} / \left(2 \left(2.6487\right)\right)$
$\implies {d}_{2} = 24.05 m$, rounded to two decimal places.

Apr 15, 2017

See below.

#### Explanation:

Calling
${v}_{i}$ the initial velocity
${v}_{1}$ the velocity just before collision
${v}_{2}$ the velocity just after collision
${d}_{2}$ the distance covered by ${m}_{2}$ after collision
$g$ the gravity acceleration (9.81 [m/s^2]), we have

Before collision

$\frac{1}{2} {m}_{1} {v}_{i}^{2} - {m}_{1} g {\mu}_{k} d = \frac{1}{2} {m}_{1} {v}_{1}^{2}$

during the collision

${m}_{1} {v}_{1} = {m}_{2} {v}_{2}$

after the collision

$\frac{1}{2} {m}_{2} {v}_{2}^{2} = {m}_{2} g {\mu}_{k} {d}_{2}$

Solving

$\left\{\begin{matrix}\frac{1}{2} {m}_{1} {v}_{i}^{2} - {m}_{1} g {\mu}_{k} d = \frac{1}{2} {m}_{1} {v}_{1}^{2} \\ {m}_{1} {v}_{1} = {m}_{2} {v}_{2} \\ \frac{1}{2} {m}_{2} {v}_{2}^{2} = {m}_{2} g {\mu}_{k} {d}_{2}\end{matrix}\right.$

for ${v}_{1} , {v}_{2} , {d}_{2}$ we obtain

$\left\{\begin{matrix}{v}_{1} = \sqrt{{v}_{i}^{2} - 2 d g {\mu}_{k}} \\ {v}_{2} = \frac{{m}_{1}}{{m}_{2}} \sqrt{{v}_{i}^{2} - 2 d g {\mu}_{k}} \\ {d}_{2} = {\left({m}_{1} / {m}_{2}\right)}^{2} \left(\frac{{v}_{i}^{2} - 2 d g {\mu}_{k}}{2 g {\mu}_{k}}\right)\end{matrix}\right.$

Putting values

${v}_{1} = 5.13 , {v}_{2} = 10.66 , {d}_{2} = 21.45$