bullet of mass 80 gram moving with initial velocity of 100
m/s strikes a wooden block and comes to rest after penetrating a
distance of 2 cm in it. Then calculate 1) Initial P, 2) Final P ,3)
Retardation caused by block 4) Resistive force exerted by block
Answers
Explanation:
mass of bullet=80gm
=80/1000kg
distance =2cm
2/100m
using Newton's third equation of motion
2as=v²-u²
2*a*2/100=0-10000
a=-25*10⁴m/s²
1. Intial momentum=80/100*100
=80kgm/s
2. object comes to rest therefore final velocity becomes zero and final p also becomes zero
3. a=-25*10⁴
4. Force=mass*acceleration
F=80/100*(-25*10⁴)
F=-2*10⁵N
Solution :
As per the given data ,
- Mass of the bullet(m) = 80 g = 0.08 kg
[ in order to convert g into kg simply divide by 1000 ]
- Initial velocity of the bullet(u) = 100 m/s
- Final velocity of the bullet (v) = 0 m/s [rest]
- Distance traveled = 2cm = 0.02 m
[ in order to convert cm into m divide by 100 ]
1) Initial momentum (Pi)
➜ Pi = m x u
➜ Pi = 0.08 x 100
➜ Pi = 8 kg m/s
2) Final momentum (Pf)
➜ Pf = m x v
➜ Pf = 0.08 x 0
➜ Pf = 0
3) Retardation caused by the block (a)
By applying third equation of motion ,
➜ v² = u² + 2as
On rearranging ,
➜ a = v ² - u² / 2s
➜ a = 0 - 10000 / 2 x 0.02
➜ a =- 10 ^6 / 4
➜ a = - 25 x 10 ^4 m/s²
4) Resistive force exerted by the block
F = m x a
F = 0.08 x - 25 x 10 ^4
F = 800 x- 25
F = - 20000 N