Question

A bullet of 10 g strikes a sand bag at a speed of 103 m/s and gets embedded after travelling 5 cm. Calculate: a. the resistive force exerted by sand on the bullet. b. the 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

2(a)(0.05)=0²-1000²2a(5/100)=-1000000

2(a)(0.05)=0²-1000²2a(5/100)=-10000002a(1/20)=-1000000

2(a)(0.05)=0²-1000²2a(5/100)=-10000002a(1/20)=-1000000a(1/10)=-1000000

2(a)(0.05)=0²-1000²2a(5/100)=-10000002a(1/20)=-1000000a(1/10)=-1000000a=-10000000=-10power7

2(a)(0.05)=0²-1000²2a(5/100)=-10000002a(1/20)=-1000000a(1/10)=-1000000a=-10000000=-10power7acceleration of bullet is -10power7or -10000000

By using F=ma

y using F=maF=0.01(10000000)

y using F=maF=0.01(10000000)F=1/100(10000000)

y using F=maF=0.01(10000000)F=1/100(10000000)F=-100000N

y using F=maF=0.01(10000000)F=1/100(10000000)F=-100000Nthere fore resistive force of bullet =-100000N

b) By using,

a=v-u/t

=v-u/t-10000000=0-1000/t

=v-u/t-10000000=0-1000/t-10000000=-1000/t

=v-u/t-10000000=0-1000/t-10000000=-1000/t-10000=t

=v-u/t-10000000=0-1000/t-10000000=-1000/t-10000=tor

=v-u/t-10000000=0-1000/t-10000000=-1000/t-10000=tort=-10⁴seconds

Hope it helps!!

Thank you ✌️

Answer 2

Answer:

Resistive Force = 1060.9 N

Time taken by bullet to come to rest = 9.7 x 10⁻⁴seconds

Step-by-step explanation:

Given,

Mass of bullet m = 10 g = 0.01 kg

Speed, v = 103 m/s

Distance, s = 5 cm = 0.05 m

We know that F = ma  

and v^2 = u^2 – 2as  

or a = (v^2 – u^2) / 2s

Now F = m (v^2 – u^2 / 2s)

= 0.01 x ((103^2 – 0) / 2 x 0.05)

Hence,

Resistive force =  F = 1060.9 N

Now time taken to come to rest=

S = (u + v / 2) t

So, t = 2s / u + v

 = 2 x 0.05 / 103

t = (9.7)^(-4) seconds

Hence,

Time taken by bullet to come to rest = 9.7 x 10⁻⁴seconds

HOPE IT HELPS,

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