A body weighing 2.0 kg moves along a horizontal path at a speed of 15 m / s. 10 s after the beginning of the action

A body weighing 2.0 kg moves along a horizontal path at a speed of 15 m / s. 10 s after the beginning of the action of a constant force, it acquires a speed of 5 m / s, directed in the opposite direction. Determine the impulse of force, the magnitude and direction of the force, the final impulse of the body

m = 2 kg.

V1 = 15 m / s.

t = 10 s.

V2 = 5 m / s.

F * t -?

R – ?

Let’s write 2 Newton’s law in vector form: m * a = F, where m is the mass of the body, a is the acceleration of the body, F is the force that acts on the body.

Let’s write the definition for acceleration in vector form: a = (V2 – V1) / t.

m * (V2 – V1) / t = F.

Let’s draw the coordinate axis in the direction of the initial motion of the body and write down Newton’s 2 law for projections onto this axis.

F = m * (- V2 – V1) / t.

F = 2 kg * (- 5 m / s – 15 m / s) / 10 s = – 4 N.

The sign “-” means that the force is directed in the opposite direction of the original movement of the body.

The impulse of force is the product of force by the time of its action.

F * t = 4 N * 10 s = 40 N * s.

p = m * V2.

p = 2 kg * 5 m / s = 10 kg * m / s.

Answer: the body was affected by a force F = 4 N directed in the opposite direction of movement, p = 10 kg * m / s.