Collision question?

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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?

2 Answers
Apr 15, 2017

Answer:

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.

Apr 15, 2017

Answer:

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#