## Question

###### Need help ASAP please A 10.2 g bullet is fired into a stationary block of wood...

need help ASAP please

A 10.2 g bullet is fired into a stationary block of wood (M= 4.60 kg). The relative motion of the bullet stops inside the block. The speed of the bullet plus wood combination is measured at 0.700 m/s. What was the original speed of the bullet? 1.16 m/s 516 m/s 310 m/s O 301 m/s 316 m/s

## Answers

A 10.2g bullet is fixed into a stationary block of wood (M = 4.60 kg) The relative motion of the bullet stops inside the block. The speed of the bullet plus wood Combination is measured at 0.700 m/s. 3 kg Given : mass of bullet (m) = 1012g = 10.2410 mass of block (M) 2 4.60 kg speed of bullet plaas block ( 0.700m/s. speed of wood block(u) = omis original speed of bullet (d) = ? law of conservation of momentum, we get mu+Mu = (1 fm) v 10.2X1830 +(4:60 x0) = (4.60+ 0.01) 0.7 Using 0.0l = 9 4.61x0.7 4.6180.7 0.01 =) Uz 322.7 mis . The osiginal speed of bullet is 322.7 m/s

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