Question

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 6.80 m before stopping. How far does the lighter fragment slide

Answer 1

Distance the lighter fragment slides before stopping= 333.2 m

Step-by-step explanation:

Gravitational acceleration = g

Mass of the lighter fragment = m

Mass of the heavier fragment = 7m

Velocity of the lighter fragment after the explosion = V1

Velocity of the heavier fragment after the explosion = V2

The object is at rest before the explosion hence the total momentum of the system is zero.

mV1 + 7mV2 = 0

V1 = -7V2

Coefficient of friction =
`\mu`

Friction force on the lighter fragment = f1 =
`\mu mg`

Friction force on the heavier fragment = f2 =
`7\mu mg`

Deceleration of the lighter fragment due to friction = a1

ma1 = -f1

ma1 =
`-\mu mg`

a1 =
`-\mu g`

Deceleration of the heavier fragment due to friction = a2

7ma2 = -f2

7ma2 = -7
`\mu mg`

a2 =
`-\mu g`

Final velocity of the lighter fragment = V3 = 0 m/s

Final velocity of the heavier fragment = V4 = 0 m/s

Distance traveled by the lighter fragment before coming to rest = d1

Distance traveled by the heavier fragment before coming to rest= d2 =6.8 m

V4² = V2² + 2a₂d₂

`,`

`(0)^2 = V_2^2 + (2)(-\mu g)d_2`

`d_2 = (V_2^2)/(2 \mu g)`

V₃² = V₁² + 2a₁d₁

`(0) = (-7V_2)^2 + 2(-\mu g)d_1`

`d_1 = (49V_2^2)/(2 \mu g)`

d₁ = 49d₂

d₁ = (49)(6.8)

d₁ = 333.2 m

Distance the lighter fragment slides before stopping = 333.2 m

Answer 2

Step-by-step 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