PHY303K HW7

# PHY303K HW7 - Perez(nap563 – HW07 – TSOI –(58160 1...

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Unformatted text preview: Perez (nap563) – HW07 – TSOI – (58160) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A uranium nucleus 238 U may stay in one piece for billions of years, but sooner or later it de- cays into an α particle of mass 6 . 64 × 10 − 27 kg and 234 Th nucleus of mass 3 . 88 × 10 − 25 kg, and the decay process itself is extremely fast (it takes about 10 − 20 s). Suppose the uranium nucleus was at rest just before the decay. If the α particle is emitted at a speed of 1 . 7 × 10 7 m / s, what would be the recoil speed of the thorium nucleus? Correct answer: 2 . 90928 × 10 5 m / s. Explanation: Let : v α = 1 . 7 × 10 7 m / s , M α = 6 . 64 × 10 − 27 kg , and M Th = 3 . 88 × 10 − 25 kg . Use momentum conservation: Before the de- cay, the Uranium nucleus had zero momentum (it was at rest), and hence the net momentum vector of the decay products should total to zero: vector P tot = M α vectorv α + M Th vectorv Th = 0 . This means that the Thorium nucleus recoils in the direction exactly opposite to that of the α particle with speed bardbl vectorv Th bardbl = bardbl vectorv α bardbl M α M Th = (1 . 7 × 10 7 m / s) (6 . 64 × 10 − 27 kg) 3 . 88 × 10 − 25 kg = 2 . 90928 × 10 5 m / s . 002 10.0 points A 6 kg steel ball strikes a wall with a speed of 10 . 4 m / s at an angle of 52 . 1 ◦ with the normal to the wall. It bounces off with the same speed and angle, as shown in the figure. x y 1 . 4 m / s 6 kg 1 . 4 m / s 6 kg 52 . 1 ◦ 52 . 1 ◦ If the ball is in contact with the wall for . 32 s, what is the magnitude of the average force exerted on the ball by the wall? Correct answer: 239 . 571 N. Explanation: Let : M = 6 kg , v = 10 . 4 m / s , and θ = 52 . 1 ◦ . The y component of the momentum is un- changed. The x component of the momentum is changed by Δ P x = − 2 M v cos θ . Therefore, using impulse formula, F = Δ P Δ t = − 2 M v cos θ Δ t = − 2 (6 kg) (10 . 4 m / s) cos 52 . 1 ◦ . 32 s bardbl vector F bardbl = 239 . 571 N . Note: The direction of the force is in negative x direction, as indicated by the minus sign. 003 10.0 points A student performs a ballistic pendulum experiment using an apparatus similar to that shown in the figure. Perez (nap563) – HW07 – TSOI – (58160) 2 Initially the bullet is fired at the block while the block is at rest (at its lowest swing point). After the bullet hits the block, the block rises to its highest position, see dashed block in the figure, and continues swinging back and forth. The following data is obtained: the maximum height the pendulum rises is 3 cm, at the maximum height the pendulum sub- tends an angle of 36 . 9 ◦ , the mass of the bullet is 99 g, and the mass of the pendulum bob is 762 g....
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PHY303K HW7 - Perez(nap563 – HW07 – TSOI –(58160 1...

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