Question

1. the magnitude of the net force exerted in the x-direction on a 2.50-kg particle varies in time as shown in figure p9.19. find (a) the impulse of the force over the 5.00-s time interval, (b) the final velocity the particle attains if it is originally at rest, (c) its final velocity if its original velocity is -2.00î m/s, and (d) the average force exerted on the particle for the time interval between 0 and 5.00 s.

figure p9.19
chart: f (n) on y-axis, t (s) on x-axis, with a red trapezoidal graph from t=0 to t=5, peaking at t=2 to t=3 with f=4n

Explanation:

Step1: Calculate impulse (area under F-t)

The impulse $J$ is the area of the triangle in Figure P9.19. Area formula: $\frac{1}{2} \times \text{base} \times \text{height}$
$$J = \frac{1}{2} \times 5.00\ \text{s} \times 4.00\ \text{N} = 10.0\ \text{kg·m/s}$$

Step2: Find final velocity (rest initial)

Use impulse-momentum theorem $J = m(v_f - v_i)$. $v_i=0$, $m=2.50\ \text{kg}$
$$v_f = \frac{J}{m} = \frac{10.0}{2.50} = 4.00\ \hat{i}\ \text{m/s}$$

Step3: Find final velocity ($v_i=-2.00\hat{i}$)

Use $J = m(v_f - v_i)$, solve for $v_f$
$$v_f = \frac{J}{m} + v_i = 4.00\ \hat{i} + (-2.00\ \hat{i}) = 2.00\ \hat{i}\ \text{m/s}$$

Step4: Calculate average force

Average force $F_{avg} = \frac{J}{\Delta t}$, $\Delta t=5.00\ \text{s}$
$$F_{avg} = \frac{10.0}{5.00} = 2.00\ \hat{i}\ \text{N}$$

Answer:

(a) $10.0\ \text{kg·m/s}$
(b) $4.00\hat{i}\ \text{m/s}$
(c) $2.00\hat{i}\ \text{m/s}$
(d) $2.00\hat{i}\ \text{N}$