Question

A bullet of mass 50g strikes a wooden block with speed 1000m/s and penetrates up to a distance of 5 cm.calculate the resistive force exerted by the wooden block on bullet and the time taken by the bullet to come to rest.​

Answer 1

Given :

• Mass of bullet, m = 50 g = 50/1000  kg = 0.05 kg
• Initial speed of bullet, u = 1000 m/s
• distance covered by bullet, s = 5 cm = 5/100  m = 0.05 m
• Final velocity of bullet, v = 0

To find :

• Resistive force exerted by wooden block on bullet, F = ?
• Time taken by the bullet to come to rest, t = ?

Knowledge required :

• Third equation of motion

$\red{\bigstar}\boxed{\sf{v^2-u^2=2\;a\;s}}$

• Newton's second Law of motion

$\red{\bigstar}\boxed{\sf{F=m\;a}}$

• First equation of motion

$\red{\bigstar}\boxed{\sf{v=u+a\;t}}$

[ where F is the force exerted, m is the mass , a is acceleration, v is final velocity, u is initial velocity, s is distance covered and t is time taken ]

Solution :

Calculating acceleration of the bullet

Using third equation of motion

$\longmapsto\sf{v^2-u^2=2\;a\;s}$

$\longmapsto\sf{(0)^2-(1000)^2=2\;a\;(0.05)}$

$\longmapsto\sf{-1000000=0.1\;a}$

$\longmapsto\underline{\underline{\red{\sf{a=-10000000\;ms^{-2}}}}}$

Calculating Force exerted by the bullet on the wooden block

Using Newton's second law of motion

$\longmapsto\sf{F=m\;a}$

$\longmapsto\sf{F=(0.05)\times(-10000000)}$

$\longmapsto\boxed{\underline{\underline{\large{\red{\sf{F=-500000\;N=-500\;kN}}}}}}$

Since, The bullet came to rest on collision with the wooden block therefore, Force exerted by Wooden block on the bullet will be equal to the force exerted by bullet on wooden block.

so,

• Magnitude of Force exerted by wooden block on bullet is 500 kilo Newton. ( In direction opposite to motion of bullet )

Calculating time taken by bullet to come to rest

Using first equation of motion

$\longmapsto\sf{v=u+a\;t}$

$\longmapsto\sf{0=1000+(-10000000)\;t}$

$\longmapsto\sf{-1000=-10000000\;\;t}$

$\longmapsto\boxed{\underline{\underline{\large{\red{\sf{t=0.0001\;sec=0.1\;milli\;sec}}}}}}$

so,

• Time taken by bullet to come to rest is 0.1 millisecond.

Answer 2

Explanation:

Given; mass is 50 g, initial velocity is 1000 m/s, final velocity is 0 m/s and distance is 5 cm or 0.05 m.

Using the third equation of motion,

v² - u² = 2as

Substitute the values,

→ (0)² - (1000)² = 2a(0.05)

→ -1000000 = 0.1a

→ a = -10000000 m/s²

Therefore, the acceleration of the bullet is -10000000 m/s². (Negative sign Retardation).

Now, using the first equation of motion,

v = u + at

Substitute the values,

→ 0 = 1000 + (-10000000)t

→ 1000/10000000 = t

→ 0.0001 = t

The time taken by the bullet is 0.0001 sec.

Also,

Force = Mass × Acceleration

Substitute the known values,

→ Force = 0.05 × (-10000000)

→ Force = -500000 N

Hence, the force is 500000N.