Question

A bullet of mass 0.2 kg is headed at horizontal speed 80 m/s toward a block of mass 12(0.2 kg) = 2.4 kg resting on a horizontal frictionless surface. The bullet strikes the block and comes to rest relative to the block somewhere deeply embedded inside the block. What average force did the block exert on the bullet while it was coming to a stop inside the block, if the stopping process took 0.06 s?

Answer 1

-246.1538 N

Step-by-step explanation:

`m_1` = Mass of bullet = 0.2 kg

`m_2` = Mass of block = 2.4 kg

`u_1` = Initial Velocity of bullet = 80 m/s

`u_2` = Initial Velocity of block = 0 m/s

v = Combined velocity of block and bullet

As linear momentum is conserved

`m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=(m_1u_1 + m_2u_2)/(m_1 + m_2)\\\Rightarrow v=(0.2* 80 + 2.4* 0)/(0.2 + 2.4)\\\Rightarrow v=6.15384\ m/s`

The velocity of the block and bullet together is 6.15384 m/s

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

Equation of motion

`v=u+at\\\Rightarrow a=(v-u)/(t)\\\Rightarrow a=(6.15384-80)/(0.06)\\\Rightarrow a=-1230.769\ m/s^2`

From Newton's second law of motion

`F=ma\\\Rightarrow F=0.2* -1230.769\\\Rightarrow F=-246.1538\ N`

The average force the block exerted on the bullet while it was coming to a stop inside the block is -246.1538 N