Question

Question :-

A bullet of 10 g strikes a sand bag at a speed of 103 ms-1 and gets embedded after travelling 5 cm. Calculate
(i) the resistive force exerted by the sand on the bullet.
(ii) the time taken by the bullet to come to rest.
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Answer 1

$\huge\underline\mathtt{Answer}$

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__________________

Answer 2

Answer:

Explanation:

Given:

Mass of bullet (m) = 10 g = 10/1000 = 0.01 kg

Initial velocity (u) = 10^3 m/s or 1000 m/s

Final velocity (v) = 0 m/s

Distance traveled by the bullet (s) = 5 cm = 5/100 = 0.05 m

To find:

(i) The resistive force exerted by the sand on the bullet (F) = ?

(ii) The time taken by the bullet to come to rest (t) = ?

Solution:

▪︎ v^2 - u^2 = 2as

$- - > \: \frac{ {v}^{2} \: - {u}^{2} }{2 \times s} = \: a$

$- - > \: \frac{ {0}^{2} - {1000}^{2} }{2 \times 0.05} = \: a$

$- - > \: \frac{ - 1000000}{0.1} = \: a$

$- - > \: { - 10000000} \: \frac{m}{ {s}^{2} } = a$

$- - > \: { - 10}^{7} \: \frac{m}{ {s}^{2} } = \: a$

(i) F = m × a

here, F = force applied to the body (SI unit = Newton)

m = mass of the body (SI unit = kg)

a = acceleration of the body (SI unit = m/s^2)

(putting values)...

==> F = 0.01 × -10^7

==> F = -10^5 N

(ii) v = u + at

here, u = initial velocity of the body (SI unit = m/s)

v = Final velocity (SI unit = m/s)

t = time of application (SI unit = seconds)

a = acceleration (SI unit = m/s^2)

(putting values)...

==> 0 = 10^3 - 10^7× t

==> -10^3 = -10^7 × t

==> 10^3/10^7= t

==> t = 10^-4 s

Hence, The resistive force = -10^5 Newtons and

Time taken = 10^-4 seconds

Hope this helped you dear...