Question

A book slides off a horizontal tabletop. As it leaves the table’s edge, the book has a horizontal velocity of magnitude v0. The book strikes the floor in time t. If the initial velocity of the book is doubled to 2v0, what happens to (a) the time the book is in the air, (b) the horizontal distance the book travels while it is in the air, and (c) the speed of the book just before it reaches the floor? In particular, does each of these quantities stay the same, double, or change in another way? Explain.

Answer 1

Answer:

(a) The time the book is in the air stays the same.

(b) The horizontal distance the book travels doubles.

(c) The speed of the book just before it reaches the floor increases but not doubles.

Explanation:

The following kinematics equations will be used to solve this question:

$v_y = v_{y_0} + a_yt\\y - y_0 = v_{y_0}t + \frac{1}{2}a_yt^2\\v_y^2 = v_{y_0}^2 + 2a_y(y - y_0)$

(a) Initially, the y-component of the velocity is zero. So, the x-component of the velocity is doubled.

We will use the second equation for both cases:

$0 - y_0 = v_{y_0}t - \frac{1}{2}(g)t^2 = 0 - \frac{1}{2}gt^2\\0 - y_0 = v_{y_0}t_2 -\frac{1}{2}gt_2^2 = 0 - \frac{1}{2}gt_2^2\\t = t_2$

Since, the initial velocity in the y-direction is zero, and the height of the table is constant. The time it takes from the edge of the table to the floor is the same.

(b) For the horizontal distance the book travels, we should use the second equation again, and keep in mind that the acceleration in the x-direction is zero.

$x - 0 = v_{x_0}t + \frac{1}{2}a_xt^2 = v_0t\\x_2 = 2v_0t_2 = 2v_0t$

Hence, the horizontal distance doubles.

(c) The vertical velocity does not change, since the initial velocity in the y-direction is zero. The horizontal velocity does not change along the motion, since the acceleration in the x-direction is zero. However, since the initial velocity in the x-direction doubles, the final velocity in the x-direction doubles as well.

$v_x = v_{x_0} + a_xt = v_{x_0}$

However, this does not mean that final speed of the book will double. Because the speed of the object is calculated as follows:

$v_{\rm final_1} = \sqrt{v_x^2 + v_y^2} = \sqrt{v_0^2 + v_y^2}\\v_{\rm final_2} = \sqrt{v_{x_2}^2 + v_y^2} = \sqrt{(2v_0)^2 + v_y^2}$

As can be seen from above, the final speed increases but not doubles.