A meteorite of mass $1500$kg moves with a speed of $0.700c$. What is the magnitude of its momentum $p$?

• A. $\frac{1500\times 0.700c}{\sqrt{1-{\frac{0.700c}{c}}^2}}$
• B. $1500\times 0.700c\times \sqrt{1-{\frac{0.700c}{c}}^2}$
• C. $\frac{1500\times 0.700c}{\sqrt{1-{\frac{0.700}{1}}^2}}$
• D. $\frac{1500\times 0.700c}{1}$

Answer: (C)

Verified step by step guidance

1. Identify the given quantities: mass of the meteorite \(m = 1500\ \text{kg}\) and its speed \(v = 0.700c\), where \(c\) is the speed of light.
2. Recall that the relativistic momentum \(p\) is given by the formula: \(p = \frac{m v}{\sqrt{1 - \left(\frac{v}{c}\right)^2}}\)
3. Substitute the given values into the formula: \(p = \frac{1500 \times 0.700c}{\sqrt{1 - (0.700)^2}}\)
4. Calculate the denominator inside the square root first: \$1 - (0.700)^2 = 1 - 0.49 = 0.51$
5. Take the square root of the denominator and then divide the numerator by this value to find the magnitude of the momentum \(p\).