Question

A truck starts moving with initial speed u and acceleration 'a' on a straight road. A ball is projected vertically with speed V from the truck at t = 0. Then -

• A. Trajectory of the ball as seen by a man standing on the road is a straight line.
• B. Trajectory of the ball as seen by a man on the truck is a straight line.
• C. The ball will land on the truck only if a = 0.
• D. The ball will land on the truck for all values of a.

Answer 1

Answer: (C)

The ball will land on the truck only if a = 0.

Answer 2

Let's analyze the motion of the ball from two different frames of reference.

1. Frame of reference: Man standing on the road (ground frame)

The truck starts with initial speed u and acceleration a. A ball is projected vertically with speed V from the truck at t = 0. At t = 0, the velocity of the ball in the ground frame has two components: Horizontal velocity: The ball shares the truck's initial horizontal velocity, which is u. Vertical velocity: The ball is projected vertically relative to the truck with speed V. Since the truck's vertical velocity is 0, the ball's initial vertical velocity in the ground frame is V.

So, the initial velocity of the ball in the ground frame is (u_x, u_y) = (u, V). The acceleration of the ball in the ground frame is (a_x, a_y) = (0, -g) (ignoring air resistance).

The position of the ball at time t in the ground frame, assuming the projection point is the origin (0, 0) at t=0, is given by: x(t) = u_x * t + 0.5 * a_x * t^2 = u * t + 0.5 * 0 * t^2 = u * t y(t) = u_y * t + 0.5 * a_y * t^2 = V * t + 0.5 * (-g) * t^2 = V * t - 0.5 * g * t^2

To find the trajectory, we eliminate t. From the horizontal equation, t = x / u (assuming u != 0). Substituting this into the vertical equation: y = V * (x / u) - 0.5 * g * (x / u)^2 y = (V/u) * x - (g / (2 * u^2)) * x^2

This equation is of the form y = Ax - Bx^2, which is the equation of a parabola. If u = 0, the trajectory is a vertical line (x=0, y = Vt - 0.5gt^2). The question states the truck "starts moving with initial speed u", implying u is the initial velocity, not necessarily zero. In the general case where u != 0, the trajectory is a parabola. Therefore, the statement "Trajectory of the ball as seen by a man standing on the road is a straight line" is false.

2. Frame of reference: Man on the truck (truck frame)

This frame is non-inertial if the truck is accelerating (a != 0). The initial velocity of the ball relative to the truck is given as purely vertical with speed V. So, (u_x', u_y') = (0, V).

The acceleration of the ball relative to the truck is a_ball_truck = a_ball_ground - a_truck_ground. a_ball_ground = (0, -g) a_truck_ground = (a, 0) a_ball_truck = (0 - a, -g - 0) = (-a, -g)

The position of the ball at time t relative to the truck, assuming the projection point is the origin (0, 0) relative to the truck at t=0, is given by: x'(t) = u_x' * t + 0.5 * a_x' * t^2 = 0 * t + 0.5 * (-a) * t^2 = -0.5 * a * t^2 y'(t) = u_y' * t + 0.5 * a_y' * t^2 = V * t + 0.5 * (-g) * t^2 = V * t - 0.5 * g * t^2

To find the trajectory in the truck frame, we eliminate t. If a = 0, then x'(t) = 0. The trajectory is described by x' = 0 and y'(t) = Vt - 0.5gt^2. This is a vertical line segment (up and down). If a != 0, we have x'(t) = -0.5 * a * t^2 and y'(t) = V * t - 0.5 * g * t^2. These are parametric equations. From the first equation, t^2 = -2x'/a. Substituting this into the second equation is not straightforward to get a simple y'(x') form, but the path is not a straight line unless a=0. The equation y' = V*sqrt(-2x'/a) + (g/a)*x' for t = sqrt(-2x'/a) shows a dependency involving a square root, which is not a straight line equation. The trajectory is a parabola in the truck frame when a != 0. Therefore, the statement "Trajectory of the ball as seen by a man on the truck is a straight line" is false in general (only true if a=0).

3. Condition for the ball to land on the truck

The ball is projected from a point on the truck. It will land on the truck if, at the moment the ball returns to its initial vertical level (y = 0 in the ground frame, or y' = 0 in the truck frame), its horizontal position is the same as the horizontal position of the point on the truck from which it was projected.

Let's use the ground frame. The time of flight T is when the ball returns to y = 0. y(T) = V * T - 0.5 * g * T^2 = 0 T (V - 0.5 * g * T) = 0 Since T > 0 for the end of the flight, T = 2V / g.

During this time T, the horizontal position of the ball is x_ball(T) = u * T = u * (2V / g). The horizontal position of the truck at time T is x_truck(T) = u * T + 0.5 * a * T^2. x_truck(T) = u * (2V / g) + 0.5 * a * (2V / g)^2 = u * (2V / g) + 2 * a * V^2 / g^2.

For the ball to land on the truck, x_ball(T) must be equal to x_truck(T). u * (2V / g) = u * (2V / g) + 2 * a * V^2 / g^2 0 = 2 * a * V^2 / g^2

Since V > 0 (projection speed) and g > 0, this equation is satisfied only if a = 0. Thus, the ball will land on the truck only if the acceleration of the truck is zero.

Let's confirm this using the truck frame. The ball lands on the truck if its position relative to the truck is (0, 0) at the time of flight T. The time of flight relative to the truck (when y'(T) = 0) is T = 2V/g, which is the same as in the ground frame. At time T, the horizontal position of the ball relative to the truck is: x'(T) = -0.5 * a * T^2 = -0.5 * a * (2V / g)^2 = -0.5 * a * (4V^2 / g^2) = -2 * a * V^2 / g^2. For the ball to land on the truck, x'(T) must be 0. -2 * a * V^2 / g^2 = 0 This implies a = 0 (since V > 0, g > 0).

Both frames of reference yield the same condition: the ball lands on the truck only if a = 0.

Let's evaluate the given options:

1. Trajectory of the ball as seen by a man standing on the road is a straight line. (False, it's a parabola for u != 0)
2. Trajectory of the ball as seen by a man on the truck is a straight line. (False, it's a parabola for a != 0)
3. The ball will land on the truck only if a = 0. (True, as derived above)
4. The ball will land on the truck for all values of a. (False, it lands only if a = 0)

Therefore, the only correct statement is that the ball will land on the truck only if a = 0.