Solve this question about impulse and momentum: One hazard of space travel is debris left by previous missions. There are several thousand masses large enough to detect by radar orbiting the earth, but there are far greater numbers of very small masses such as flakes of paint. Calculate the force exerted by a 0.190 mg chip of paint that strikes a space shuttle window at a relative speed of 5.00 ✕ 103 m/s and sticks, given the collision lasts 6.00 ✕ 10-8 s. Such a collision chipped the window of the ill-fated Challenger in June 1983, causing $50,000 of damage
First, we need to calculate the momentum of the chip of paint before and after the collision.
The momentum before the collision can be calculated using the equation:
\[p_{\text{before}} = m \cdot v\]
Where:
m = mass = 0.190 mg = 0.190 x 10^-6 kg
v = velocity = 5.00 x 10^3 m/s
\(p_{\text{before}} = (0.190 \times 10^{-6} \, \text{kg}) \times (5.00 \times 10^{3} \, \text{m/s})\)
\(p_{\text{before}} = 9.50 \times 10^{-4} \, \text{kg m/s}\)
Since the chip sticks to the window, the momentum after the collision is equal to the momentum before the collision:
\(p_{\text{after}} = 9.50 \times 10^{-4} \, \text{kg m/s}\)
Next, we can calculate the force exerted by the chip using the impulse-momentum theorem:
\[F \cdot t = \Delta p\]
Where:
F = force
t = time = 6.00 x 10^-8 s
\(\Delta p = p_{\text{after}} - p_{\text{before}}\)
Substitute the values we have:
\[F \cdot 6.00 \times 10^{-8} = 9.50 \times 10^{-4} - 9.50 \times 10^{-4}\]
\[F = \frac{9.50 \times 10^{-4}}{6.00 \times 10^{-8}}\]
\[F ≈ 158.33 N\]
Therefore, the force exerted by the 0.190 mg chip of paint that strikes a space shuttle window at a relative speed of 5.00 x 10^3 m/s and sticks is approximately 158.33 N.