6 kg block slides to the east across a frictionless surface with a momentum of 30 kg x meters per second strikes an object with an impulse of 10 newtons per second to the west of the block, what is the speed of the block at collission

Explanation:

Initial momentum = 30 kg m/s .

impulse on the object = 10 N s towards west . = - 10 Ns

impulse = change in momentum

Let velocity after impulse be v

momentum after impulse = 6 v

change in momentum = 6v - 30

So

- 10 = 6v - 30

20 = 6v

v = 3.33 m /s

Since the velocity is positive , that means it is moving in the same direction as earlier .

82. Since you have a distance and a force, then the easiest principle to use is energy, i.e. work.

The work done by friction is F * d. This work cancels out the kinetic energy of the stone (1/2)mv^2

Fd = (1/2)mv^2

F = (1/2)mv^2/d.

Plug in m = 20 kg, v = 3 m/sec, d = 40 m.

83. With more mass, the kinetic energy is higher now. The work needed is higher. W = F * d and F is the same.

Explanation:Hope I helped :)