Question

7. A bullet of mass 0.04kg moving with a speed of 90ms' enters a heavy wooden block and
is stopped after a distance of 60cm. what is the average resistive force excerted on the
bullet?
1
a. 200N
b. 250N
c. 260N
d. 270N
​

Answer 1

✯ Average resistive force excerted on the bullet = 270 N ✯

Explanation:

Given:

• Mass of bullet (m) = 0.04 kg
• Initial velocity (u) = 90 m/s
• Distance travelled (s) = 60 cm

To find:

• Average resistive force excerted on the bullet.

Solution:

The bullet is stopped after a distance of 60 cm.

So,

Final velocity (v) will be 0 m/s.

• Distance (s) = 60 cm = 0.6 m

We know that,

${\boxed{\sf{v^2=u^2+2as}}}$

• [ Put values]

$\implies$ 0² = 90² + 2a × 0.6

$\implies$ 1.2a +8100 = 0

$\implies$ 1.2a = -8100

$\implies$ a = -8100/1.2

$\implies$ a = - 6750

Therefore, the acceleration of the bullet is - 6750 m/s².

We know that,

${\boxed{\sf{F=ma}}}$

• [put values]

$\implies$ F = 0.04×-6750

$\implies$ F = -270

† “ - ” sign indicates resistive force which is opposing.

Therefore,

Option (d) 270 N is correct.

______________

Answer 2

Given :

• mass = m= 0.04kg
• initial velocity = u=90m/s
• final velocity = v= 0m/s
• displacement = s= 60cm =0.6m

━━━━━━━━━━━━━━━━━

Need to Find:

• The Force acting on body

━━━━━━━━━━━━━━━━

Solution :

By 3rd equation of motion we know,

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

$\implies \: {0}^{2} - {90}^{2} = 2a \times 0.6$

$\implies \: - 8100 = 0.6 \times 2 \times a$

$\implies \: a = \dfrac{ - 81000}{2 \times 6}$

$\purple{\bold{\boxed{ \implies a= 6750 m/ s^{2}}}}$

━━━━━━━━━━━━━━━━━

We know,

F= m× a

⟹ F= 0.04 × (-6750) N

⟹ $\red{\bold{\boxed{\large{Force= (-270)N}}}}$

The Force of 270N is acting in opposite to the direction of motion.