Show units and work for each part A 11-g bullet with an initial speed of 290 m/s
ID: 1371371 • Letter: S
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Show units and work for each part
A 11-g bullet with an initial speed of 290 m/s is shot directly at a 1.1-kg wooden block, which rests on a frictionless surface. What is the speed of the block, after the bullet is embedded in the block? (Although this is not justified, calculate your answer to 4 significant figures. This will uncover arty conceptual errors you might make.) A 11-g bullet with an unknown initial speed is shot directly at a 1-kg wooden block, which rests on a frictionless surface. After impact, me bullet is embedded Into the block, and the block moves at a speed of 2.92 m/s. What was the initial speed of the bullet? (Although this is not justified. calculate your answer to 4 significant figures. This will uncover any conceptual errors you might make.) A 10-g bullet with an initial speed of 303 m/s is shot directly at a 1-kg wooden block, which rests on a frictionless surface. The bullet passes through the block and emerges with a speed of 270 m/s. What is the speed of the block, after the bullet has passed through the block? D. A 10-g bullet with an initial speed of 301 m/s is shot directly at a 1-kg wooden block, which rests on a frictionless surface. The bullet passes through the block, after which the block is moving at a speed of 0.186 m/s. What is the speed of the bullet after it has passed through the block? E. ? A 11-g bullet with an unknown initial speed is shot directly at a 1-kg wooden block, which rests on a frictionless surface. The bullet passes through the Sock and emerges with a speed of 305 m/s, after which the block is moving at a speed of 0.185 m/s. What was the speed of the bullet before it hit the block?Explanation / Answer
(A) Speed of the block, after the bullet is embedded in the block whih will be given as :
using conservation of momentum, we have
mb v0,b + mw vo,w = (mb + mw) vw { eq.1 }
where, mb = mass of bullet = 11 g = 0.011 kg
v0,b = initial speed of bullet = 290 m/s
mw = mass of wooden block = 1.1 kg
vo,w = initial speed of wooden block = 0 m/s
inserting the values in above eq.
(0.011 kg) (290 m/s) + (1.1 kg) (0 m/s) = [(0.011 kg) + (1.1 kg)] vw
(3.19 kg.m/s) = (1.11 kg) vw
vw = (3.19 kg.m/s) / (1.11 kg)
vw = 2.87 m/s
(b) The initial speed of bullet which will be given as :
using conservation of momentum, we have
mb v0,b + mw vo,w = (mb + mw) vw
where, vw = speed of wooden block after impact = 2.92 m/s
mw = mass of wooden block = 1 kg
vo,w = initial speed of wooden block = 0 m/s
mb = mass of bullet = 11 g = 0.011 kg
inserting these values in above eq.
(0.011 kg) v0,b + (1 kg) (0 m/s) = [(0.011 kg) + (1 kg)] (2.92 m/s)
(0.011 kg) v0,b = (2.95 kg.m/s)
v0,b = (2.95 kg.m/s) / (0.011 kg)
v0,b = 268.1 m/s
(c) The speed of the block, after the bullet has passed through the block which will be given as :
using conservation of momentum, we have
mb v0,b + mw vo,w = mb vb + mw vw { eq.2 }
where, mb = mass of bullet = 10 g = 0.01 kg
v0,b = initial speed of bullet = 303 m/s
mw = mass of wooden block = 1 kg
vo,w = initial speed of wooden block = 0 m/s
vb = final speed of the ball = 270 m/s
inserting these values in eq.2,
(0.01 kg) (303 m/s) + (1 kg) (0 m/s) = (0.01 kg) (270 m/s) + (1 kg) vw
(3.03 kg.m/s) = (2.7 kg.m/s) + (1 kg) vw
(0.33 kg.m/s) = (1 kg) vw
vw = 0.33 m/s
(d) The speed of bullet after it has passed through the block which will be given as :
using conservation of momentum, we have
mb v0,b + mw vo,w = mb vb + mw vw { eq.2 }
where, mb = mass of bullet = 10 g = 0.01 kg
v0,b = initial speed of bullet = 301 m/s
mw = mass of wooden block = 1 kg
vo,w = initial speed of wooden block = 0 m/s
vw = final speed of the wooden block = 0.186 m/s
inserting these values in eq.2,
(0.01 kg) (301 m/s) + (1 kg) (0 m/s) = (0.01 kg) vb + (1 kg) (0.186 m/s)
(3.01 kg.m/s) = (0.01 kg) vb + (0.186 kg.m/s)
(2.82 kg.m/s) = (0.01 kg) vb
vb = 282 m/s
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