Question

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​

Answer 1

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

Answer 2

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