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A bullet of mass m hits an obstacle and is deflected through 60 deg from its initial line of motion, while its speed changes from u to v. What is the magnitude of the impulse delivered to the bullet?

1. m*sqrt(u² + v² - u*v)
2. m*sqrt(u² + v² + u*v)
3. m*sqrt(u² + v² - 2*u*v)
4. m*(u + v)

Correct answer: m*sqrt(u² + v² - u*v)

Solution

The magnitude of the change in momentum is found from the triangle of initial and final momentum vectors separated by 60 deg, giving J = m*sqrt(u² + v² - 2uv*cos60) = m*sqrt(u² + v² - uv).

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