Chapter 3, Problem 83A

a.

To determine

Would a ball dropped by an astronaut hit the surface of the moon with greater, equal or lesser speed than that of a ball dropped from the same height on the earth.

Answer to Problem 83A

Lesser Speed

Explanation of Solution

Given information:The value of the free fall acceleration on the moon is about one sixth of its value on earth.

Formula Used: $v^2=u^2+2ah$

Here v is final velocity, u is initial velocity and a is free fall acceleration and h is height.

Calculation:

Free fall acceleration on the earth is denoted by g.

Let the height through which ball falls is h .

For free fall initial velocity u is zero

So from equation of motion

  $v^2=u^2+2ah$

Here v is final velocity, u is initial velocity and a is free fall acceleration and h is height.

$v^2=0+2\times \frac{g}{6}\times h$   …(1)

For earth, this equation is $v^2=0+2gh$   …(2)

Comparing equation 1 and 2, v is less in equation 1.

Conclusion:

So, a ball dropped by an astronaut hits the surface of the moon with lesser speed than that of a ball dropped from the same height on the earth.

b.

To determine

Whether it take the ball to more, less or equal time to fall.

Answer to Problem 83A

More Time

Explanation of Solution

Given information:The value of the free fall acceleration on the moon is about one sixth of its value on earth.

Formula Used: $h=ut+\frac{1}{2}g{t}^2$

Calculation: A stopped car is an example of zero acceleration. although normal forces and force due to gravity works on it. The resultant of these forces is zero.

From equation of motion, $h=ut+\frac{1}{2}g{t}^2$

Let at the surface of the moon time is $t_1$ and on earth time taken is $t_2$ .

So, at the surface of moon, $h=0+\frac{1}{2}\frac{g}{6}{t}_1$ Or $t_1=\frac{12h}{g}$   ...(1)

And at the surface of earth, $h=0+\frac{1}{2}g{t}_2$ Or $t_2=\frac{2h}{g}$   ...(2)

From equation 1 and 2, $t_1$ is greater than $t_2$ .

Conclusion:

So, ball will take more time to fall on the surface of the moon.