Question

A bullet of mass 0.0021 kg moving at 404 m/s
impacts a large fixed block of wood and travels
6.5 cm before coming to rest.
Assuming that the deceleration of the bullet
is constant, find the magnitude exerted by the
wood on the bullet.
Answer in units of kN.

Answer 1

$\huge{\red{\mathfrak{\underline{Happy\:Diwali}}}}$

✨✨Here is your solution

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$since \: deceleration \: is \: constant \: so \\ {v}^{2} - {u}^{2} = - 2as \\ {0}^{2} - {(404)}^{2} = - 2a \: \frac{6.5}{100} \\ a = \frac{ {(404)}^{2} \times 100}{2 \times 6.5} \\ \\ force = m \times a \\ = 0.0021 \times \frac{ {(404)}^{2} \times 100}{2 \times 6.5} \\ = \: \frac{21 \times {10}^{ - 4} \times {10}^{2} \times {(404)}^{2} \: }{2 \times 6.5} \\ 12555.076 \times {10}^{ - 2} \\ = 0.12 \times {10}^{5} \times {10}^{ - 2} \\ 0.12kN \:$

Answer 2

In first, we have to find the deceleration of the bullet (a) by using a linear motion equation. Which is,

 $V^{2} = U^{2} + 2aS$

Where, V = final velocity of the bullet = 0

           U = initial velocity of the bullet = 404 m/s

          a = acceleration of the bullet = -a

           S = distance traveled by bullet = 6.5 cm = 0.065 m

$0 = 404^{2} - 2a*0.065\\ a = 163216/0.13\\ a = 1255507.7 m/s^{2}$

Now we can find force by using Newton's second law,

     $F= ma\\ F = 0.0021*1255507.7=2636.6 N = 2.64 kN$

Answer : 2.64 kN