Question

c) A bullet having a mass of 10 g and moving with a speed of 1.5 m/s, penetrates a thick wooden plank of mass 900 g. The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity.


d) A person swims 100 m in the first 40 s, 80 m in the next 40 s and 45 min the last 20 s. What is the average speed?

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Answer 1

Answer:

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1)

Mass of bullet= 10g= 0.01kg

Velocity of bullet= 1.5m/s

Mass of plank= 900g=0.09kg

Total momentum before collision= (mass of plank) x (velocity)

= (0.01x1.5) + (0.09x0)=0.015 kg m/s

Total momentum after collision= (mass of bullet +mass of plank) x (velocity)

=(0.1 kg x V) =0.1 V kg m/s

According to law of conservation of momentum,

Total momentum before collision=Total momentum after collision

m1v1 = m2V2

0.015kg m/s =0.1 v kg m/s

Thus, velocity acquired by the plank and the bullet = 0.15m/s

2)

Total distance = 100+ 80+45 m = 225m

Total time = 40+40+20s = 100s

Average speed = total distance /total time

Average speed = 225m/100s

Average speed = 2.25m/s

Hope this helps uh

Answer 2

Answer:

Answer:

$\displaystyle\huge\red{\underline{\underline{AnSwEr}}}$

AnSwEr

1)

Mass of bullet= 10g= 0.01kg

Velocity of bullet= 1.5m/s

Mass of plank= 900g=0.09kg

Total momentum before collision= (mass of plank) x (velocity)

= (0.01x1.5) + (0.09x0)=0.015 kg m/s

Total momentum after collision= (mass of bullet +mass of plank) x (velocity)

=(0.1 kg x V) =0.1 V kg m/s

According to law of conservation of momentum,

Total momentum before collision=Total momentum after collision

m1v1 = m2V2

0.015kg m/s =0.1 v kg m/s

Thus, velocity acquired by the plank and the bullet = 0.15m/s

2)

Total distance = 100+ 80+45 m = 225m

Total time = 40+40+20s = 100s

Average speed = total distance /total time

Average speed = 225m/100s

Average speed = 2.25m/s

Hope this helps uh