Question

You drop a 14-g ball from a height of 1.5 m and it only bounces back to a height of 0.85 m. what was the total impulse on the ball when it hit the floor? (ignore air resistance.)

Answer 1

Answer:

The impulse on the ball when it hit the floor is equal to 0.133 kg-m/s.

Explanation:

Given that,

Mass of the ball, m = 14 g = 0.014 kg

It is dropped from a height of 1.5 m and it only bounces back to a height of 0.85 m. The conservation of energy is applied here.

$\dfrac{1}{2}mv^2=mgh$

If h = 1.5 m

$v=\sqrt{2gh}$

$v_1=\sqrt{2\times 9.8\times 1.5}$

$v_1=5.42\ m/s$

If h = 0.85 m

$v_2=\sqrt{2\times 9.8\times 0.85}$

$v_2=-4.08\ m/s$ (as the ball bounces back so it is negative)

The impulse on the ball when it hit the floor is equal to the change in momentum of the ball as :

$J=m(v_2-v_1)$

$J=0.014\times (-4.08-5.42)$

J = -0.133 kg-m/s

So, the impulse on the ball when it hit the floor is equal to 0.133 kg-m/s. Hence, this is the required solution.

Answer 2

Remember that  impulse = change in momentum 

this means we compute the momentum of the ball just before impression and just after; we know the mass, so we find the speeds 

the ball falls for 1.5m and will achieve a speed given by energy conservation: 

1/2 mv^2 = mgh => v=sqrt[2gh]=5.42m/s 

since it rises only to 0.85 m, we compute the initial speed after power from the same equation and get 
v(after)=sqrt[2*9.81m/s/s*0.85m] = 4.0837 m... 

now, recall that momentum is a vector, so that the momentum down has one sign and the momentum up has a positive sign, so we have 

impulse = delta (mv) = m delta v = 0.014 kx (4.08m/s - (-5.42m/s) = 0.133 kgm/s