A 5 gram bullet moving with an initial speed of 400 m/sis fired into AND PASSES
ID: 1723840 • Letter: A
Question
A 5 gram bullet moving with an initial speed of 400 m/sis fired into AND PASSES THROUGH a 1.5kg block. The block isinitially at rest on a frictionless, horizontal surface, and isconnected to a spring with force constant 800 N/m. The BLOCK moves5cm to the right after impact. A) What is the intial speed of the block just as the bulletemeges from it (hint: think mechanical energy)? B) What is the speed at which the bullet emerges from theblock (hint: think momentum)? C) What is the mechanical energy converted into internalenergy in the collision? Thanks! A 5 gram bullet moving with an initial speed of 400 m/sis fired into AND PASSES THROUGH a 1.5kg block. The block isinitially at rest on a frictionless, horizontal surface, and isconnected to a spring with force constant 800 N/m. The BLOCK moves5cm to the right after impact. A) What is the intial speed of the block just as the bulletemeges from it (hint: think mechanical energy)? B) What is the speed at which the bullet emerges from theblock (hint: think momentum)? C) What is the mechanical energy converted into internalenergy in the collision? Thanks!Explanation / Answer
Given that the mass of bullet is m = 5 gm= 0.005 kg Initial speed of bullet is u1 = 400m/s mass of block is M = 1.5 kg initial speed of bullet is u2 = 0 m/s spring constant is k = 800 N /m Compressed length is x = 0.05 m ---------------------------------------------------------------- Let final velocity of the bullet v1 and final velocity of theblock is v2 (a) Apply conservation of energy just after collision and atcompressed length. (1/2)M v22 = (1/2) kx2 v22 = k x2 / M v2 = [ k x2 / M ]1/2 =1.154 m/s (b) Apply conservation of momentum before and aftercollision m*u1 + M*u2 = m*v1 + M*v2 m*u1 + 0 = m*v1 + M*v2 ( since u2 =0 m/s ) v1 = ( m*u1 - M*v2 ) / m = ----------m/s (c) The mechanical energy ( kinetic energy ) of blockconverted into internal energy.
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