Question #9cedf

1 Answer
Aug 3, 2016

Answer:

#2/5#

Explanation:

The idea here is that the ratio of the ball's velocity when it first makes contact with the ground, and its velocity when it leaves the ground does not depend on the value of #g#, which is why the problem doesn't provide it in the first place.

Let's take #h_1# to be the initial height of the ball and #h_2# the height it reaches after it bounces. Also, let's say that #v_1# is the velocity of the ball when it first hits the ground and #v_2# is the velocity with which it leaves the ground.

The ball is dropping from #h#, which means that you can write

#v_1^2 = overbrace(color(white)(a)v_0^2color(white)(a))^(color(blue)(=0)) + 2 * g * h_1#

#v_1^2 = 2 * g * h_1" " " "color(orange)((1))#

After it hits the ground, it bounces with an initial velocity of #v_2# and reaches the height #h_2# when its velocity is once again equal to zero. You thus have

#overbrace(color(white)(a)v_f^2color(white)(a))^(color(blue)(=0)) = v_2^2 - 2 * g * h_2#

#v_2^2 = 2 * g * h_2" " " "color(orange)((2))#

Now, in order to find the factor by which the ball loses its velocity, divide equation #color(orange)((2))# by equation #color(orange)((1))#

#v_2^2/v_1^2 = (color(red)(cancel(color(black)(2 * g))) * h_2)/(color(red)(cancel(color(black)(2 * g))) * h_1)#

This gets you

#v_2/v_1 = sqrt(h_2/h_1)#

Plug in your values to find

#v_2/v_1 = sqrt(( 1.8 color(red)(cancel(color(black)("m"))))/(5color(red)(cancel(color(black)("m"))))) = sqrt(36/100) = 6/10 = 3/5#

You can thus say that

#v_2 = 3/5 * v_1#

Sine the velocity of the ball decreased by

#v_1 - v_2 = v_1 - 3/5 * v_1 = (1 - 3/5) * v_1 = 2/5 * v_1#

you can say that it decreased by a factor of

#(v_1 - v_2)/v_1 = (2/5 * color(red)(cancel(color(black)(v_1))))/color(red)(cancel(color(black)(v_1))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2/5)color(white)(a/a)|)))#