Question

A 3.00-kg object is initially moving northward at 15.0 m/s. Then a force of 15.0 N, toward the east, acts on it for 4.40 s.A. At the end of the 4.40 s, what is magnitude of the object’s final velocity?B. What is the direction of the final velocity? Enter the angle in degrees where positive indicates north of east and negative indicates south of east.C. What is the change in momentum during the 4.40 s? Take +y to be north and +x to be east. Enter a positive answer if the change in momentum is toward east and a negative answer if the change in momentum is toward west.

Answer 1

ANSWERS

A) 26.6 m/s

B) 34°

C) 66 kg·m/s

Step-by-step explanation

A) Let's draw a diagram of this situation to understand it better

The force is horizontal, so we have Fx = 15N. By Newton's second law of motion:

`F_x=a_x\cdot m`

The horizontal acceleration of the object is

The initial horizontal velocity of the object was 0m/s - because it was moving vertically. Therefore, the final horizontal velocity of the object after applying the force for 4.4 seconds is:

`\begin{gathered} v_x=u_x+a_xt \\ v_x=0+5m/s^2\cdot4.4s \\ v_x=22m/s \end{gathered}`

But the object had an initial velocity in the north direction of vy = 15m/s, which is unaffected because the force is applied horizontally. Now we have the two components of the velocity

Note that they form a right triangle where the vector is the hypotenuse and the components are the legs. Hence, we can find the magnitude by the Pythagorean Theorem:

`v=\sqrt[]{v^2_x+v^2_y}=\sqrt[]{22^2+15^2}=\sqrt[]{709}\approx26.6m/s`

B) As shown in the explanation of item A), we have a triangle formed by the velocity vector and its components. To find the direction we have to find the angle θ. Since it's a right triangle we can use the tangent of the angle:

`\tan \theta=(v_y)/(v_x)`

Solving for θ:

`\theta=\tan ^(-1)(15)/(22)=36.287\approx34`

So the direction of the final velocity is 34° north of east (so it is positive).

C) Change of momentum is:

`\Delta p=m(v_f-v_i)`

Since we are working with two dimensional motion, we have to analyze each coordinate separately and then find the magnitude and direction.

However, in the vertical direction there's no change of momentum because there's no external impulse in that direction - in other words, the momentum is conserved in the vertical direction.

Since the force is applied in the horizontal direction, the object has a change of momentum in that direction.

The equation above is correct, but also momentum is related to force applied by the equation

`F=(\Delta p)/(\Delta t)`

We'll use this equation instead, because for this one we have to use the given information and not the calculated one - such as velocity. Solving for momentum

`\Delta p=F\cdot\Delta t=15N\cdot4.4s=66kg(m)/(s)`