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A bullet of mass 50 g moving with an initial velocity of 100 m s-1, strikes a wooden block and comes to rest after penetrating a distance 2 cm in it. Calculate (i) initial momentum of the bullet (ii) final momentum of the bullet, (iii) retardation caused by the wooden block, and (iv) resistive force exerted by the wooden block

Laws of Motion

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Answer

Given,

m = 50 g

Converting g to Kg

1000 g = 1 Kg

50 g = (11000\dfrac{1}{1000}) x 50 kg

50 g = 0.05 Kg

Hence, m = 0.05 Kg

u = 100 m s-1

S = 2 cm

Converting cm to m

100 cm = 1 m

2 cm = (1100\dfrac{1}{100}) x 2 = 0.02 m

(i) Initial momentum of the bullet = mass (m) x initial velocity (u)

Substituting the values in the formula above we get,

Initial momentum = 0.05 x 100 = 5 kg m s-1

Hence, Initial momentum = 5 kg m s-1

(ii) Final momentum = mass (m) x final velocity (u)

Substituting the values in the formula above we get,

Final momentum = 0.05 x 0 = 0 kg m s-1

Hence, Final momentum = Zero

(iii) As we know,

v2 - u2 = 2as

Substituting the values in the formula above we get,

021002=2×a×0.02-10000=0.04×aa=100000.04a=10000004a=250000m s20^2 - 100^2 = 2 \times a \times 0.02 \\[0.5em] - 10000 = 0.04 \times a \\[0.5em] \Rightarrow a = - \dfrac {10000}{0.04} \\[0.5em] \Rightarrow a = - \dfrac {1000000}{4} \\[0.5em] \Rightarrow a = - 250000 \text{m s}^{-2}\\[0.5em]

Hence,

Retardation caused by wooden block = 2.5 x 105 m s-2

(iv) Now,

Force (f) = mass (m) x acceleration (a)

Substituting the values in the formula above we get,

F=0.05×(2.5×105)F=0.125×105NF = 0.05 \times (2.5 \times 10^5) \\[0.5em] \Rightarrow F = 0.125 \times 10^5 \text{N} \\[0.5em]

Hence,

Resistive force exerted by wooden block = 12500 N

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