Question

A bullet of 10kg strikes a sand bag at a speed of 1000m/s and gets embedded after travelling 5 cm.calculate (a) Resistive force exerted by sand on bullet (b )time taken by bullet to come to rest

Answer 1

mass of bullet should be =10gm=0.01kg
initial velocity (U)of bullet=1000m/s
final velocity (V)=0m/s(because it has stopped after striking sand)
displacement of bullet=5cm=0.05m
a)using 2as=V ²-U²
2(a)(0.05)=0²-1000²
2a(5/100)=-1000000
2a(1/20)=-1000000
a(1/10)=-1000000
a=-10000000=-10power7
acceleration of bullet is -10power7or -10000000
by using F=ma
F=0.01(10000000)
F=1/100(10000000)
F=-100000N
there fore resistive force of bullet =-100000N

b)by using a=v-u/t
-10000000=0-1000/t
-10000000=-1000/t
-10000=t
or
t=-10⁴seconds

Answer 2

Answer:

(i) F = 10^5N

(ii) t = 10^-4s

Explanation:

Answer:

Given

m = 10 g = 10/1000 kg

u = = 103 m/s

v = 0

s = 5 cm = 5/100 m

(i) the resistive force exerted by the sand on the bullet

By using the formula

v2 – u2= 2as

0 – (103)2 = 2 × a × 5/100

– (103)2 = 10/100 a

– (103)2 = 0.1 a

a = -106/0.1

a = -107ms-2

F = m × a

F = (10/1000) × -107

F = 10^5N

(ii) the time taken by the bullet to come to rest.

v = u + at

0 = 103 + (-107) × t

107t = 103

t = 103/107

t = 10^-4s

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