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 6.10 m before stopping. How far does the lighter fragment slide

298.9m

Explanation:

Given that : Note that the object exploded into two.

Let mass of lighter fragment be m

Let mass of heavier fragment = 7m

Let the velocity of lighter fragment = V1

Let the velocity of heavier fragment = V2

Recall that object was at rest before explosion occur. Hence, momentum of system is zero.

mv1 + 7mv2 = 0

V1 = -7V2

force of Friction on lighter fragment

f1 = μmg

force of friction on heavier fragment

f2 = 7μmg

Deceleration of lighter fragment = a1

we know that force of friction is given as

ma1 = - f1

ma1 = -μmg

a1 = -μmg/m

a1= -μg

Deceleration of heavier fragment = a2

7ma2 = -f2

7ma2 = - μ7mg

a2 = -μg

Let the final velocity of lighter fragment = V3 = 0 m/s

Let the final velocity of heavier fragment = V4 = 0 m/s

Distance traveled by lighter fragment before rest position is = d1

Distance traveled by the heavier fragment before rest position is = d2 = 6.10 m

From :

V^2 = u^2 + 2as

V4^2 = v2^2+2a2d2

d2 = v2^2/2μg ............ (1)

V3^2 = v1^2+2a1d1

0 = (-7v2)^2+2(-μg)d1

0 = 49v2^2+2(-μg)d1

Make d1 the subject

0 = 49v2^2-2(μg)d1

2(μg)d1 = 49v2^2

d1 = 49v2^2/2μg............ (2)

Put equation (1) into (2)

d1 = 49d2

d1 = 49×6.1

d1 = 298.9m