ment FULL SCREEN PRINTER VERSION ·BACK NEXT MESSAGE MY INSTRUCTOR Chapter 09, Pr
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ment FULL SCREEN PRINTER VERSION ·BACK NEXT MESSAGE MY INSTRUCTOR Chapter 09, Problem 066 Block 1, with mass m? and speed 5.1 m/s, slides along an x axis on a frictionless floor and then undergoes a one- dimensional elastic collision with stationary block 2, with mass m2-0.64m1. The two blocks then slide into a region slide? where the coefficient of kinetic friction is 0.47; there they stop. How far into that region do (a) block 1 and (b) block 2 (a) Number Units Units (b) Number Show Work is REQUIRED for this question: Open Show Work SHOW HINT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO HINI-LECTURE LINK TO TEXT LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE By accessing this Question Assistance, you will learn while ydu earn points based on instructor the Point Potential Policy set by your SAVE FOR LATER SUBMIT ANSWER Question Attempts: o of 3 usedExplanation / Answer
ELASTIC COLLISION
mass of block1 = m1 = m mass of block2 ,m2 = 0.64*m1 = 0.64*m
speeds before collision
v1i = 5.1 m/s v2i = 0 m/s
speeds after collision
v1f = ? v2f = ?
initial momentum before collision
Pi = m1*v1i + m2*v2i
after collision final momentum
Pf = m1*v1f + m2*v2f
from momentum conservation
total momentum is conserved
Pf = Pi
m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)
from energy conservation
total kinetic energy before collision = total kinetic energy after collision
KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2
KEf = 0.5*m1*v1f^2 + 0.5*m2*v2f^2
KEi = KEf
0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)
solving 1&2
we get
v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)
v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)
final speed of block1
v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)
v1f = ((m-0.64m)*5.1 + (2*0.64m*0))/(m + 0.64m )
v1f = 1.12 m/s
v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)
v2f = ((0.64m-m)*0 + (2*m*5.1))/( m + 0.64m )
v2f = 6.22 m/s
In the region
frictional force fk = -uk*m*g
work done = W*d = -uk*m*g*d
from work energy relation
work = change in KE = K2 - K1
K2 = 0
K1 = (1/2)*m*vf^2
-uk*m*g*d = -(1/2)*m*vf^2
distance d = vf^2/(2*uk*g)
for block1
d1 = v1f^2/(2*uk*g)
d1 = 1.12^2/(2*0.47*9.8)
d1 = 0.136 m <<<<-----------ANSWER
for block 2
d2 = v2f^2/(2*uk*g)
d2 = 6.22^2/(2*0.47*9.8)
d2 = 4.2 m <<<<-----------ANSWER
DONE please check the answer. any doubts post in comment box
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