Answer:
Explanation:
Given that,
One fragment is 7 times heavier than the other
Let one fragment mass be M
Let this has a velocity v
And the other 7M
And this a velocity V
Initially the fragment is at rest u = 0
Applying conservation of momentum
Momentum is given as p=mv
Initial momentum = final momentum
Po = Pf
(M+7M) × 0 = 7M •V − Mv
0 = 7M•V - Mv
Divide both sides by M
0 = 7V -v
v = 7V
Since friction decelerates the masses to zero speed, we can calculate the NET work on the individual blocks and relate this quantity to the change in kinetic energy of each block
The workdone by the 7M mass is
Distance moved by 7M mass is 6.8m, Then, d =6.8m
W = fr × d
Where fr = µkN
When N=W =mg, where m=7M
N= 7Mg
fr = −µk × 7mg
Then, W(7m) = −7µk•Mg×d
W(7m) = −7µk•Mg×6.8
W(7m) = −47.6 µk•Mg
Then, same procedure,
Let distance move by the small mass be m
Work done by M mass
W(m) = −µk•Mg×d'
Since it is a wordone by friction, that is why we have a negative sign.
Using conservation of energy
Work done by 7M mass is equal to work done by M mass
W(7m) = W(m)
−47.6 µk•Mg = −µk•Mg×d
Then, M, g and µk cancels out
We are left with
-46.7 = -d
Then, d = 46.7m
