 # Selina Solutions Class 9 Physics Solutions for Exercise 3C in Chapter 3 - Chapter 3 Laws Of Motion

A bullet of mass 50g moving with an initial velocity of 100m/s, strikes a wooden block and comes to rest after penetrating a distance 2cm 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

Given: m=50g or 0.05kg, u=100m/s, v=0m/s, s=2cm or 0.02m

(i) Initial momentum of the bullet = mu

= 0.05 x 100 = 5m/s

(ii) Final momentum of the bullet = mv

= 0.05 x 0 = 0 m/s

(iii) Retardation caused by the wooden block

Retardation is negative acceleration.

We know from the equation of motion that,

\begin{array}{l} \mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{as} \\ 0^{2}-(100)^{2}=2 \mathrm{a}(0.02) \\ (100 \times 100) / 0.04=0.04 \mathrm{a} \\ \quad \mathrm{a}=-2.5 \times 10^{5} \mathrm{m} / \mathrm{s}^{2} \end{array}

(iv) Force = mass x acceleration

\begin{array}{l} =0.05 \times\left(2.5 \times 10^{5}\right) \\ =12500 \mathrm{N} \end{array}

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