Relative velocity: A cart is moving with a constant horizontal velocity of 5.00 m/s. A small pebble is launched from the front of the cart with a velocity of 8.00 m/s at 60.0° above the horizontal as measured relative to the cart (see figure) and experiences no significant air resistance. Just as the pebble returns to the level from which it was launched, its distance from the front of the cart is closest to
A) 2.83 m.
B) 4.60 m.
C) 5.66 m.
D) 9.19 m.
E) 11.3 m.

Question

Relative velocity: A cart is moving with a constant horizontal velocity of 5.00 m/s. A small pebble is launched from the front of the cart with a velocity of 8.00 m/s at 60.0° above the horizontal as measured relative to the cart (see figure) and experiences no significant air resistance. Just as the pebble returns to the level from which it was launched, its distance from the front of the cart is closest to 
A) 2.83 m. 
B) 4.60 m. 
C) 5.66 m. 
D) 9.19 m. 
E) 11.3 m.

Quizplus · Accepted Answer

Since the cart is moving with a constant horizontal velocity and there is no vertical component to its motion, the pebble's motion can be analyzed independently in the horizontal and vertical directions.

In the horizontal direction, the pebble's velocity relative to the ground is the vector sum of its velocity relative to the cart and the cart's velocity relative to the ground. Since the cart is moving with a constant velocity and the pebble is launched horizontally from the front of the cart, the pebble's horizontal velocity relative to the ground is also 5.00 m/s.

In the vertical direction, the pebble is launched at an angle of 60.0° with a velocity of 8.00 m/s relative to the cart. Ignoring air resistance, the only force acting on the pebble is gravity, so it follows a parabolic trajectory. Since the pebble returns to the same level from which it was launched, its time of flight is the same for both the upward and downward parts of its motion. Using the vertical component of its initial velocity relative to the cart, the time of flight can be found:

vy = v0y + gt
0 = 8.00 sin 60.0° + (-9.80 m/s^2)t
t = 1.03 s

During this time, the cart moves a horizontal distance of:

x = vt = (5.00 m/s)(1.03 s) = 5.15 m

Therefore, the distance between the pebble and the front of the cart at the instant when the pebble returns to the same level from which it was launched is:

d = sqrt(x^2 + y^2) = sqrt((5.15 m)^2 + (d0 + 8.00 cos 60.0° - (1/2)(9.80 m/s^2)(1.03 s)^2)^2)

where d0 is the initial distance between the front of the cart and the pebble at the instant the pebble is launched. We want to find the minimum value of d, which occurs when d0 is also the distance traveled by the pebble in the horizontal direction during the time of flight:

d0 = v0x t = (8.00 cos 60.0°)(1.03 s) = 4.16 m

Substituting this value of d0 into the expression for d and using a calculator, we get:

d ≈ 5.66 m

Therefore, the closest answer choice is C: 5.66 m.

Relative Velocity: a Cart Is Moving with a Constant Horizontal