Question

A bullet of mass 0.05kg moving with an initial velocity 100m/s, strikes a wooden block and comes to rest after penetrating a distance 0.02m in it. (i) Initial momentum of the bullet is: a. 0.5 kgm/sb. 10 kgm/sc. 5 kgm/sd. 15 kgm/s(ii)Final momentum of the bullet bea. 0 kgm/sb. 10 kgm/sc. 5 kgm/sd. 15 kgm/s(iii)Retardation caused by the wooden block is: a. 12.5 x 105m/s2 b. 2.5 x 105 m/s2 c. 25 x 105m/s2 d. 10 x 105 m/s2(iv)Resistive force exerted by the wooden block is: a. 25000Nb. 50000Nc. 12500Nd. 6250N(v)The unit of linear momentum is: a. Kg m/s2 b. N.sc. Kg2 m/sd. None of the above

Answer 1

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Answer 2

( i )  - c. 5 kgm/sd

( ii )  - a. 0 kgm/sb

( iii )  - b. 2.5 ×10⁵ $ms^-^2$

( iv )  - c. 12500N

( v )  - d. None of the above

Given,

Mass (m) = 50 gm = 0.05 kg

Initial velocity (u) = 100 m/s

Final velocity (v) = 0

Distance (s) = 2cm = 0.02 m

(i) Initial momentum = mu

                                 = (0.05) x (100)

                                 = 5 kg m/s

(ii) Final momentum = mv

                                 = (0.05) x (0)

                                 = 0 kg m/s.

(iii) Acceleration (a) = $\frac{v^2-u^2}{2s}$

                             a = $\frac{0^2-100^2}{2*0.02}$

                             a = -250000

                                = -2.5×105 m/s²

Therefore, retardation is 2.5 ×10⁵ $ms^-^2$

(iv) Resistive Force, F = ma

F = 0.05×2.5×105

F = 12500 N

(v) The unit of Linear momentum  

       =  $Kg m/s$

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