A 3.0-kg body is initially moving northward at \(15 \mathrm{m} / \mathrm{s}\) Then a force of \(15 \mathrm{N},\) toward the east, acts on it for 4.0 s. (a) At the end of the 4.0 s, what is the body's final velocity? (b) What is the change in momentum during the \(4.0 \mathrm{s} ?\)

The initial conditions are given in the problem. The body has a mass of 3.0 kg and is initially moving northward at 15 m/s. There is also an eastward force of 15 N acting on the body.

As the force is acting horizontally (eastward), we can use Newton's second law of motion to find the acceleration. The equation is \(F = ma\), where F represents force (15 N), m is the mass (3.0 kg), and a is the acceleration in the eastward direction. So, we can calculate acceleration as \(a = \frac{F}{m}\). Calculating the acceleration: \( a = \frac{15 N}{3.0 kg} = 5\, m/s^2\) (eastward)

The acceleration in the eastward direction, along with the time interval (4.0 seconds), will affect the horizontal velocity, while the initial velocity affects the vertical direction. Use this formula to find the final horizontal velocity, \(v_x = a_x\cdot t \), and the final vertical velocity, \(v_y = v_{0y}\) (since there's no vertical acceleration): Calculating the final horizontal velocity: \( v_x = 5\, m/s^2 \times 4.0\, sec = 20\, m/s \) (eastward) The final vertical velocity will stay constant at: \(v_y = 15 \,m/s \)

Now we have the final horizontal velocity and the vertical velocity. We can use the Pythagorean theorem to find the magnitude of the final velocity vector \(v_{final}\): \(\sqrt{v_x^2 + v_y^2}\) Calculating the final velocity: \(v_{final} = \sqrt{(20 \,m/s)^2 + (15\, m/s)^2} = \sqrt{625} = 25\, m/s\)

The initial momentum \(p_{initial}\) is given by \(m \cdot v_{initial}\), where \(v_{initial}\) is the initial velocity (northward). The final momentum \(p_{final}\) is given by \(m \cdot v_{final}\). Calculating the initial and final momenta: \(p_{initial}= 3.0\, kg \times 15\, m/s = 45\, kg\cdot m/s^{(north)}\) \(p_{final}= 3.0\, kg \times 25\, m/s = 75\, kg\cdot m/s\) Now we can find the change in momentum \(\Delta p = p_{final} - p_{initial}\). Since the initial and final velocities have different directions, the change in momentum also has a direction. We will calculate the change in momentum in the horizontal and vertical directions separately. Change in horizontal momentum: \(\Delta p_x = 3.0\, kg \cdot \left( 20\, m/s\right) - 0 = 60\, kg\cdot m/s^{(east)}\) Change in vertical momentum: \(\Delta p_y = 3.0\, kg \cdot \left( 15\, m/s\right) - 3.0\, kg \cdot \left( 15\, m/s\right) = 0\, kg\cdot m/s\) To find the magnitude of the change in momentum, use the Pythagorean theorem: \(\Delta p = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2}\) Calculating the change in momentum: \(\Delta p = \sqrt{(60\, kg\cdot m/s)^2 + (0\, kg\cdot m/s)^2} = 60\, kg\cdot m/s\) (a) The final velocity of the body after 4.0 seconds is \(25 \,m/s\). (b) The change in momentum during the 4.0-second interval is \(60\, kg\cdot m/s\).