Question

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?

Answer 1

I get
A. `11.29ms^-1`
B. `24.05m`
Results rounded to two decimal places.

Explanation:

A. Initial velocity of block of mass `m_1=6.25ms^-1`
During sliding movement impeding force of friction
`F=mu_kR`
where `R` is normal reaction and is `=mg`
`:.` Force of friction `F=mu_km_1g`
Deceleration produced due to this force`=F/m_1=(mu_km_1g)/m_1=mu_kg`
`a=-0.27xx9.81=-2.6487ms^-2` .....(1)

using the kinematic expression
`v^2-u^2=2as` ......(2)
`v^2-(6.5)^2=2(-2.6487)2.4`
`v=5.43ms^-1`
Momentum of mass `m_1` just before collision`=m_1v=74.37kgms^-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 `=74.37kgms^-1`
Velocity of second mass `="momentum"/m_2=74.37/6.5=11.29ms^-1`

B. Using (1) for Deceleration due friction also for block `m_2`, and using kinematic expression (2) we get
`0^2-(11.29)^2=2(-2.6487)d_2`
`=>d_2=(11.29)^2/(2(2.6487))`
`=>d_2=24.05m`, rounded to two decimal places.

Answer 2

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

`1/2m_1v_i^2-m_1g mu_k d = 1/2m_1 v_1^2`

during the collision

`m_1 v_1 = m_2v_2`

after the collision

`1/2m_2 v_2^2=m_2g mu_k d_2`

Solving

`{(1/2m_1v_i^2-m_1g mu_k d = 1/2m_1 v_1^2),(m_1 v_1 = m_2v_2),(1/2m_2 v_2^2=m_2g mu_k d_2):}`

for `v_1,v_2,d_2` we obtain

`{(v_1 =sqrt[v_i^2-2 d g mu_k] ),(v_2=(m_1)/(m_2)sqrt[v_i^2-2 d g mu_k]),(d_2=(m_1/m_2)^2((v_i^2-2 d g mu_k)/(2gmu_k))):}`

Putting values

`v_1 = 5.13, v_2 = 10.66, d_2=21.45`