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

445.9 m

Explanation:

The law of conservation posits that the total momentum of two objects prior to collision is equal to the total momentum of the two objects after the collision has occurred.

i.e

where;

m = mass of the object

v = velocity of the object

here, the initial momentum

The final momentum = sum of the momentum of heavier fragment and lighter fragment.

=     ----- expression (1)

where

the mass and the velocity of the lighter fragment

= the mass and the velocity of the heavier fragment

From the question; we can say:

  (i.e the one that is seven times as massive as the other)

we write the above expression (1) as:

Since:

0 =

0 =

Let's determine the kinetic energy and the work done by the frictional force in both fragments

To start with the lighter fragment; the kinetic energy is expressed as;

where;

m = mass

v = v₁ = velocity

= change in direction

So;

       --------- equation (1)

For the heavier fragment ;

   ------------ equation (2)

Equating equation (1) by equation (2); we have:

Replacing ; from the above relation of the velocities; we have

∴ The lighter fragment slide 445.9 m far from the initial point of explosion.

c, 1a hope this

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s