Question

Find the resistance force offered by a wood which stops a bullet of mass 50 g

moving with a velocity of 150 m/s in a time interval of 0.005 s.​

Answer 1

Explanation:

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A bullet of mass 50 g moving with an initial velocity of 100ms

1

strikes a wooden block and comes to rest after penetrating a distance of 2cm in it calculate

(i) initial momentum of a bullet

(ii) final momentum of the bullet

(iii) retardation caused by the wooden block and

(iv) resistive force exerted by the wooden block

Hard

Solution

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Mass of the bullet is m

b

=50gm=0.05Kg

Initial velocity of bullet u

b

=100m/s

Final velocity is v

b

=0m/s

(1) Initial momentum of bullet p

i

=m

b

u

b

=5Kgm/s

(2) Final momentum of bullet p

f

=m

b

v

b

=0.05×0=0

(3) Retardation occurred by the wooden block is

v

2

−u

2

=2as

−(100)

2

=2a(0.02)

a=−250000m/s

2

(4) Resistive force

F=ma

=0.05×−250000

=12500N

Answer 2

Given: A wood stops a bullet of mass 50 g moving with a velocity of 150 m/s in a time interval of 0.005 s.​

To find: The resistance force offered by the wood.

Solution:

• Force is given by the formula,

       $F = m * a$

• Here, F is the resistance force of the wood, m is the mass of the bullet and a is the acceleration produced.
• Acceleration of the bullet is given by the formula,

        $a = \frac{v}{t}$

• Here, v is the velocity of the bullet and t is the time taken.

        $a = \frac{150ms^{-1} }{0.005s}$

           $= 30000 ms^{-2}$

• Now, the force is calculated as,

        $F = 0.05kg * 30000ms^{-2}$

            $= 1500 N$

Therefore, the resistance force offered by the wood is 1500 N.