Question

A bullet of mass 20g is horizontally fired with a horizontal velocity 150ms‐¹ from a pistol of mass 2kg. What is the recoil velocity of the pistol?​

Answer 1

Explanation:

$\Large\underline{\underline{\sf{\purple{Required\ Answer:-}}}}$

➽ For bullet :-

$\sf \: \: \: \: \: \: \: \: \: {m_1\ =\ 20g\ =\ 0.02kg,}$

$\sf \: \: \: \: \: \: \: \: \: {u_1\ =\ 0ms^{-1},\ v_1\ =\ +150ms^{-1}}$

[By convention, the direction of bullet is taken from left to right (positive)]

➽ For pistol :-

$\sf \: \: \: \: \: \: \: \: \: {m_2\ =\ 2kg,\ u_2\ =\ 0ms^{-1},}$

➴ Total momenta of the pistol and bullet before the fire,

$\sf \: \: \: \: \: \: \: \: \: {=\ m_1 u_1\ +\ m_2 u_2}$

$\sf \: \: \: \: \: \: \: \: \: {=\ 0.02kg \times 0ms^{-1}\ +\ 2kg \times 0ms^{-1}}$

$\sf \: \: \: \: \: \: \: \: \: {=\ 0kgms^{-1}}$

➴ Total momenta of the pistol and bullet after the fire,

$\sf \: \: \: \: \: \: \: \: \: {=\ m_1 v_1\ +\ m_2 v_2}$

$\sf \: \: \: \: \: \: \: \: \: {=\ 0.02kg \times (+150ms^{-1})\ +\ 2kg\ +\ v_2\ ms^{-1}}$

$\sf \: \: \: \: \: \: \: \: \: {=\ (3\ +\ 2v_2)kgms^{-1}}$

✪ According to the law of conservation of momentum ✪

$\sf \odot\ {Total\ momenta\ after\ the\ fire\ =\ Total\ momenta\ before\ the\ fire}$

$\sf \: \: \: \: \: \: \: \: \: {=\ 3\ +\ 2v_2\ =\ 0}$

$\sf \: \: \: \: \: \: \: \: \: {=\ 2v_2\ =\ -3}$

$\sf \: \: \: \: \: \: \: \: \: {\red{=\ v_2\ =\ -1.5ms^{-1}}}$

➥ Negative sign indicates that the direction in which the pistol would recoil in opposite to that of bullet (right to left).

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★ More to know :-

✪ Conservation of momentum ✪

➽ According to the second law of motion,

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {F\ =\ \dfrac{dP}{dt}}$

If external applied force F in zero (F = 0),

$\sf {Then,}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {F\ =\ \dfrac{dP}{dt}\ =\ 0}$

$\sf {i.e.,}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {P\ =\ Constant}$

➥ In the absence of ext. applied force net linear momentum of a system is always constant.

Answer 2

Answer:

Initial momentum of the system is 0 as it is at rest.

After firing-

MV = mv

⟹V = mv/M = 0.02 × 150/2

⟹V = 1.5m/s