Question

A ball of 0.1kg is thrown straight up into the air with

Answer 1

a) The momentum of the ball at the maximum height is zero

b) The momentum of the ball at halfway is 1.06 kg m/s

Explanation:

a)

There is only one force acting on the ball during its motion: the force of gravity, acting downward. As a result, the ball hasa constant acceleration downward ($g=9.8 m/s^2$, acceleration of gravity), and therefore it is in free fall motion.

This means that the velocity of the ball is given by the equation:

$v=u-gt$

where

v is the velocity at time t

u = 15 m/s is the initial velocity

$g=9.8 m/s^2$

As we see from the equation, the term $-gt$ is negative: this means that as the ball goes higher and higher, its velocity decreases. The point of maximum height is reached when the velocity has become zero (after that point, the velocity changes direction and points downward), so when

$v=0$

The momentum of the ball at any point is given by

$p=mv$

where m = 0.1 kg is its mass. Therefore, at the maximum height, since $v=0$, the momentum is also zero:

$p=0$

b)

We can use the following suvat equation to find the maximum height reached by the ball:

$v^2-u^2=2as$

where

v = 0 is the velocity at the point of maximum height

u = 15 m/s is the initial velocity

$a=-g=-9.8 m/s^2$ is the acceleration (downward)

s is the vertical displacement (the maximum height)

Solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{0-(15)^2}{2(-9.8)}=11.48 m$

This means that the ball is halfway when its height is

$h'=\frac{s}{2}=\frac{11.48}{2}=5.74 m$

We can find the velocity of the ball v' when it is at h' by using again the same equation:

$v'^2-u^2=2ah'\\v=\sqrt{u^2+2ah'}=\sqrt{(15)^2+2(-9.8)(5.74)}=10.6 m/s$

And therefore, the momentum of the ball here is

$p'=mv'=(0.1)(10.6)=1.06 kg m/s$

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