Question

14. A bullet of mass 10 g travelling horizontally with a velocity of 150 ms strikes a
stationary wooden block and comes to rest in 0.03 s. Calculate the distance of
penetration of the bullet into the block. Also calculate the magnitude of the force
exerted by the wooden block on the bullet.​

Answer 1

ANSWER:

EXPLANATION:

INITIAL VELOCITY= 150

MASS = = 10g

FINAL VELOCITY = 0/= (SINCE BULLET COMES TO REST, v= 0 )

TIME TAKEN TO REACH v = 0.03 secs

NOW, ACCELERATION OF THIS BULLET

$= v - u \div t$

$acc = (0 - 150) \div 0.03$

$acc = - 150 \div 0.03$

$acc = - 15000 \div 3$

$acc = - 5000m \div s$

THEREFORE ACCELERATION OF THE BULLET= 5000 m/s (-) INDICATES THE OPPOSITE DIRECTION.

NOW, ACCORDING TO THIRD EQUATION OF MOTION: SQUARE OF FINAL VELOCITY=SUM OF SQUARE OF INITIAL VELOCITY AND DOUBLE OF PRODUCT OF ACCELERATION AND DISTANCE OR DISPLACEMENT.

=>>>>

${v}^{2} = {u}^{2} + 2as$

AFTER INPUTTING THE KEYWORDS IN FORMULA WE GET:

$0 = {150}^{2} + 2(5000)(s)$

$0 = 22500 + 10000s$

AFTER SHIFTING 22500 TO L. H. S WE GET

$22500 = 10000s$

(-) IS NOT ADDED BECAUSE DISTANCE IS NOT NEGATIVE.

AFTER SHIFTING 10000 TO L. H. S WE GET:

$22500 \div 10000 = s$

$s = 2.25m$

THEREFORE THE DISTANCE TILL WHICH BULLET GOT EMBEDDED IN WOODEN BLOCK IS 2.25m.

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NOW WE KNOW THAT FORCE APPLIED ON A BODY IS EQUAL TO PRODUCT OF ITS ACCELERATION AND MASS.

F= mass × acceleration

IN THIS CASE:

F=?

M=10g = 0.01kg

Ac=-5000m/s^2

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THEREFORE THE FORCE EXERTED BY WOODEN BLOCK ON BULLET IS

$f = m \times a \\ f = 0.01 \times 5000 \\ f = 50newton$

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BY SCIVIBHANSHU

THANK YOU

STAY CURIOUS

Answer 2

ur answer is attached.......