In a test for a Springfield rifle, a bullet of 9.7 g hits a block of wood of 3 k

Question

In a test for a Springfield rifle, a bullet of 9.7 g hits a block of wood of 3 kg. The block f wood is held by pendulum of length 1.5 m and the angle with respect to the vertical where the velocity goes to zero is 54 degrees. What is the speed of the bullet before impact? Assume the acceleration due to gravity to be 9.8 m/s^2. Also assume that the mass of the bullet is negligible compared to the mass of the block. 1. 1376.7 2. 1112.0 3. 1000.5 4. 1314.2 5. 1076.7 6. 858.3 7. 1035.1 8. 1522.0 9. 1187.5 10. 680.95 Answer in units of m/s.

Explanation / Answer

Here, velocity of bullet + wood after impact = sqrt(2gh)

                                                                       = sqrt[2 * 9.8 * (1.5 - 1.5cos54)]

                                                                      = 3.481 m/sec

Applying conservation of momentum

=> 9.7 * 10-3 * v + 0 = ( 3 + 9.7 * 10-3) * 3.481

=>   v =   1076.7 m/sec

=> speed of bullet before impact = 1076.7 m/sec

=>   option 5) is correct .

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