Question

A bullet of mass 25 g moving with a
velocity of 200 cm/s is stopped
within 5 cm of the target. The
average resistance offered by the
target is​

Answer 1

Required Answer:-

Here average Resistance offered by the target means the reaction force applied on the bullet which will be equal to the action force as per Newton's third law of motion.

GiveN:

First finding the accleration because we need both mass and accleration for finding the force:

• v = 0 m/s
• u = 2 m/s
• s = 0.05 m

By using 3rd equation of motion,

➻ v² = u² + 2as

➻ a = v² - u² / 2s

➻ a = 0² - 2² / 2 × 0.05 m/s²

➻ a = -4/0.1 m/s²

➻ a = -40 m/s²

Now finding the force:

By using F = ma formula,

➻ Average force = 0.025 kg × -40 m/s²

➻ Average force = -1 kg m/s² or -1 N

Hence:

The average force or resistance offered by the target is:

$\huge{ \boxed{ \sf{ \blue{-1 \: N}}}}$

1 N is the magnitude of the force. Negative sign indicates force applied on the opposite direction.

Answer 2

Answer:

Given :-

• Mass of Ball = 25 g
• Initial velocity = 2 m/s
• Distance = 0.05 m
• Final velocity = 0 m/s

To Find

Average resistance

Solution :-

By using third equation of motion.

$\sf v² = u² + 2as$

$\sf \: a = \dfrac{v {}^{2} - u {}^{2} }{2s}$

$\sf \: a = \dfrac{0 {}^{2} - {2}^{2} }{2} \times 0.05$

$\sf \: a =\dfrac{ - 4}{0.01}$

$\sf \: a = -40 m/s²$

Now,

Let's find average force applied

F = ma

$\sf \: force = 0.025 kg × -40 m/s²$

$\huge \bf \: force \: = -1 \: N$