Question

Consider a particle with initial velocity v that has magnitude 12.0m/s and is directed 60.0 degrees above the negative x axis.

a. What is the x component v_x of v?

b. What is the y component v_y of v?

c. Look at this applet (I don’t have a copy of the applet, sorry). The motion diagram for a projectile is displayed, as are the motion diagrams for each component. The x component motion diagram is what you would get if you shined a spotlight down on the particle as it moved and recorded the motion of its shadow. Similarly, if you shined a spotlight to the left and recorded the particle’s shadow, you would get the motion diagram for its y component. How would you describe the two motion diagrams for the components?

1. Both the vertical and horizontal components exhibit motion with constant nonzero acceleration.
2. The vertical component exhibits motion with constant nonzero acceleration, whereas the horizontal component exhibits constant-velocity motion.
3. The vertical component exhibits constant-velocity motion, whereas the horizontal component exhibits motion with constant nonzero acceleration.
4. Both the vertical and horizontal components exhibit motion with constant velocity.

d. How long t_s does it take for the balls to reach the ground? Use 10 m/s² for the magnitude of the acceleration due to gravity.

e. Imagine that the ball on the left is given a nonzero initial velocity in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed v_x must the ball on the left start with so that it hits the ground at the same position as the ball on the right?

 

Answer 1

a.

The direction of the particle (60 degrees above the negative x-axis) indicates that it is moving left. The hypotenuse of our triangle is the magnitude of the velocity vector (the other 2 sides are the x and y components). To find the x component of velocity, we can solve for the cosine (remember SOHCAHTOA) of 60 degrees:

cos(θ) = adjacent / hypotenuse
cos(60) = v_x / 12.0m/s
0.5 = v_x /12.0m/s

Since the particle is moving left, the answer will be negative (we could have also converted the angle to 120 degrees):

v_x = -6.00m/s

b.

Similar to Part A. Just use the sine here (opposite over adjacent). The direction is above the x-axis so the answer will be positive this time:

v_y = 10.4m/s

c.

As you can see, the two components of the motion obey their own independent kinematic laws. For the vertical component, there is an acceleration downward with magnitude \(g=10 \mathrm{~m} / \mathrm{s}^{2}\). Thus, you can calculate the vertical position of the particle at any time using the standard kinematic equation \(y=y_{0}+v_{0} t+(1 / 2) a t^{2} .\) Similarly, there is no acceleration in the horizontal direction, so the horizontal position of the particle is given by the standard kinematic equation \(x=x_{0}+v_{0} t\).
so, The answer is The vertical component exhibits motion with constant nonzero acceleration, whereas the horizontal component exhibits constant-velocity motion.

d.

The height is 5m (according to the applet which isn’t copied here and we can use the kinematic equation:

x-x₀ = v₀t + 1/2at²

Change the above into the y dimension (formula doesn’t change, just the variable names):

y-y₀ = v₀t + 1/2at²

The final y position will be -5m (5 meters below where the balls started from) and we can solve accordingly:

(-5m)-(0m) = 0(t) + 1/2(-10m/s²)t²
-5m = (-5m/s²)t²
1s² = t²

t = 1.0s

e.

The ball to the right (according to the applet) is at x = 3. So we know that the ball on the left has to move 3m to the right in the same amount of time it takes to fall to the ground (1 second from Part D, above). The problem doesn’t give any air resistance or acceleration in the x-direction, so the x velocity will be constant. We can solve this using the same equation from Part D:

x-x₀ = v₀t + 1/2at²

(3)-(0) = v₀(1) + 1/2(0)(1)²

3 = v₀(1)

v₀ = 3.0m/s