Physics Question

Question

A force of magnitude F_x acting in the x-direction on a 2.60-kg particle varies in time as shown in the figure below. (Indicate the direction with the sign of your answer.)

A coordinate plane has a horizontal axis labeled t (s) and a vertical axis labeled F_x (N). A straight line is drawn from (0,0) to (2,4), then from (2,4) to (3,4), and from (3,4) to (5,0).

(a) Find the impulse of the force. (Give your answer to one decimal place.)

 kg · m/s

(b) Find the final velocity of the particle if it is initially at rest.

 m/s

(c) Find the final velocity of the particle if it is initially moving along the x-axis with a velocity of −1.90 m/s.

 m/s

Answer 1

The impulse is just the area under the F vs. t graph. If you look at the graph, it can be split up into two triangles and a rectangle. For the first triangle, the base goes from 0 to 2 sec and the height is measured from 0 N to 4 N. Since the area of a triangle is just 1/2 * base * height, the area of the triangle is 1/2 * 4 * 2 = 4 Ns. This is the same as the triangle that goes from 3 to 5 sec and 0 to 4 N. The last piece is the rectangle in the middle. The base is from 2 to 3 sec and the height is from 0 to 4 N, giving an area of 4 Ns. Adding all of these areas gives 12 Ns.

The impulse is therefore, I = 12 Ns = 12.0 kg m/s

The impulse is equal to the change in momentum, I = Δp = mv_f - mv_i

In part a, since v_i = 0, we have mv_f = 12 kg m/s. Solving for v_f = 12.0/2.60 = 4.6 m/s

In part b, it is initially traveling in the -x direction with a speed of 1.90 m/s. So, mv_f - (-mv_i) = mv_f + mv_i=12.0

Solving for v_f = 2.7 m/s