#### Answer

(a) The change in momentum of the car due to the fly is $-0.003~kg~m/s$
(b) The change in momentum of the fly due to the car is $0.003~kg~m/s$
(c) A total of 101,010 flies are required to reduce the car's speed by 1 km/h.

#### Work Step by Step

(a) We can convert the ball's speed from units of km/h to units of m/s:
$100~km/h \times \frac{1000~m}{1~km} \times \frac{1~hr}{3600~s} = 27.777778~m/s$
By conservation of momentum, the final momentum of the system is equal to the initial momentum. We can find the final speed $v_f$ of the car and fly:
$m_f~v_f = m_0~v_0$
$v_f = \frac{m_0~v_0}{m_f}$
$v_f = \frac{(1000~kg)(27.777778~m/s)}{1000~kg+0.0001~kg}$
$v_f = 27.777775~m/s$
We can find the change in momentum of the car:
$\Delta p = p_f-p_0$
$\Delta p = m~v_f-m~v_0$
$\Delta p = m~(v_f-v_0)$
$\Delta p = (1000~kg)~(27.777775~m/s-27.777778~m/s)$
$\Delta p = -0.003~kg~m/s$
The change in momentum of the car due to the fly is $-0.003~kg~m/s$
(b) We can find the change in momentum of the fly:
$\Delta p = p_f-p_0$
$\Delta p = m~v_f-m~v_0$
$\Delta p = m~(v_f-v_0)$
$\Delta p = (0.0001~kg)~(27.777775~m/s-0~m/s)$
$\Delta p = 0.003~kg~m/s$
The change in momentum of the fly due to the car is $0.003~kg~m/s$
(c) We can find the number of flies required to reduce the car's speed by 1%:
$m_f~v_f = m_0~v_0$
$m_f~(0.99~v_0) = m_0~v_0$
$m_f = \frac{m_0}{0.99}$
$m_f = \frac{1000~kg}{0.99}$
$m_f = 1010.1010~kg$
To reduce the car's speed by 1%, the final mass of the car and the flies must be $1010.1010~kg$. Then the mass of the flies must be $10.1010~kg$. We can find the number of flies required:
$\frac{10.1010~kg}{0.0001~kg} = 101,010~flies$
A total of 101,010 flies are required to reduce the car's speed by 1 km/h.