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 8.50m before stopping. How far does the lighter fragment slide?

Assume that both fragments have the same coefficient of kinetic friction.

Answer 1

Using the law of conservation of momentum,

MV = M₁(V₁) + M₂(V₂)

where

M = initial mass of the object
V = initial velocity of the object = 0 (at rest)
M₁ = mass of the first fragment
V₁ = velocity of the first fragment after explosion
M₂ = mass of the second fragment
V₂ = velocity of the second fragment after explosion

0 = M₁(V₁) + M₂(V₂)
After explosion the mass of the first body becomes 7 times than that of second one, so

M1 = 7(M2), then the above becomes

7(M₂)(V₁) + M₂(V₂) = 0

"M₂" will cancel out, hence

7V₁ = -V₂ ------------------ 1.

From this equation, it is noted that the two fragments went into opposite directions after the explosion.

The next working equation is

Vf² - Vi² = 2as

where

Vf = final velocity of the fragment = 0 (since it will stop)
Vi = velocity of the fragment after explosion
a = acceleration of the fragment
s = distance travelled by the fragment before stopping

For the first fragment,

0 - (V₁)² = 2(a)(8.5)

-(V₁)² = 17a -- ----------- 2

From Newton's 2nd Law of Motion,

F = ma

where

F = frictional force acting on the fragment

F = mg

and therefore

mg = ma

and solving for "a",

a = g = 9.8 and substituting this into Equation 2,

-(V₁)² = 17(9.8) ---------- 3

For the lighter fragment,

0 - (V₂)² = 2(a)(S)

where

a = acceleration of the lighter fragment
S = stopping distance of the lighter fragment

As derived above,

a = g

and the above becomes,

-(V₂)² = (2)()(9.8)S

-(V₂)² = 19.6()S

Since V₂ = -7V₁ (from Equation 1), then the above becomes

-(7V₁)² = 19.6()S --------------- 4

Dividing Equation 4 by Equation 3,

-(7V₁)²/-(V₁)² = 19.6()S/17(9.8)

49 = 19.6S/17*9.8

and solving for "S"

S = 416.5 m