Consider the video you just watched. Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How far behind the cart will the ball land, compared to the distance in the original experiment? a) half as far.
b) twice as far.
c) four times as far.
d) the same distance.
e) by a factor not listed above.

b

Explanation:

Given:

- The ball is fired at a upward initial speed v_yi = 2*v

- The ball in first experiment was fired at upward initial speed v_yi = v

- The ball in first experiment was as at position behind cart = x_1

Find:

How far behind the cart will the ball land, compared to the distance in the original experiment?

Solution:

- Assuming the ball fired follows a projectile path. We will calculate the time it takes for the ball to reach maximum height y. Using first equation of motion:

                                      v_yf = v_yi + a*t

Where, a = -9.81 m/s^2 acceleration due to gravity

            v_y,f = 0 m/s max height for both cases:

For experiment 1 case:

                                     0 = v - 9.81*t_1

                                      t_1 = v / 9.81

For experiment 2 case:

                                     0 = 2*v - 9.81*t_2

                                      t_2 = 2*v / 9.81

The total time for the journey is twice that of t for both cases:

For experiment 1 case:

                                     T_1 = 2*t_1

                                     T_1 = 2*v / 9.81

For experiment 2 case:

                                     T_2 = 2*t_2

                                     T_2 = 4*v / 9.81

- Now use 2nd equation of motion in horizontal direction for both cases:

                                     x = v_xi*T

For experiment 1 case:

                                     x_1 = v_x1*T_1

                                    x_1 = v_x1*2*v / 9.81

For experiment 2 case:

                                     x_2 =  v_x2*T_2

                                    x_2 = v_x2*4*v / 9.81

- Now the x component of the velocity for each case depends on the horizontal speed of the cart just before launching the ball. Using conservation of momentum we see that both v_x2 = v_x1 after launch. Since the masses of both ball and cart remains the same.

- Hence; take ratio of two distances x_1 and x_2:

                        x_2 / x_2 = v_x2*4*v / 9.81 * 9.81 / v_x1*2*v

Simplify:

                        x_1 / x_2 = 2  

- Hence, the amount of distance traveled behind the cart in experiment 2 would be twice that of that in experiment 1.      

c. magnitude and direction