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Answer 1

Answer

Magnitude of the force = 2.5 N

Given

During the sports day celebrations of a school , a game of shooting is held . The students have to target the inflated balloons fixed at a certain distance . It is observed that each student when fires a bullet from the gun it is pushed back due to some force .

To Find

Explain the reason for this observation . Also find the magnitude of the force on the gun if the masses of the bullet and the gun are 50 g and 2 kg respectively and the bullet is fired with a velocity of 100 m/s . The bullet takes 5 ms to move through the barrel of the gun .

Concept Used

• Newton Third Law
• Conservation of momentum
• Equation of motion
• Newton Second Law

Solution

1 .

The reason for the gun to move backwards when fired a bullet from gun is because of the gun exerts force on the bullet in the forward direction , at the same time bullet also exerts force on the gun in backward direction . So the gun recoils . This whole works on Newton's 3rd law .

It states that , action and reaction are both same , but opposite in direction .

2 .

Magnitude of the force on the gun :

Apply conservation of momentum ,

$\rm \to (m_g+m_b)v_i=m_g.v_r+m_b.v_b\\\\\rm \to (m_g+m_b)(0)=2(v_r)+(0.05)(100)\ \; \\\\\rm Since,v_i=0\ m/s\ ,m_b=5\ g=0.05\ kg\\\\\rm \to 0=2v_r+5\\\\\rm \to 2v_r=-5\\\\\rm \to v_r=-2.5\ m/s\ \; \pink{\bigstar}$

Note : Negative sign of recoil velocity velocity denotes opposite direction

Initial velocity , u = Vr = -2.5 m/s

Final velocity , v = 0 m/s

Finally goes to rest

Acceleration , a = ? m/s²

Displacement , s = - 5 m

Since moves through negative direction

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (0)² - (-2.5)² = 2a(-5)

⇒ 0 - 12.5 = -10a

⇒ 10a = 12.5

⇒ a = 1.25 m/s²

Apply formula for force ,

⇒ F = ma

⇒ F = (2)(1.25)

⇒ F = 2.5 N

Magnitude of the force on the gun is 2.5 N

Answer 2

Answer:

Given

During the sports day celebrations of a school , a game of shooting is held . The students have to target the inflated balloons fixed at a certain distance . It is observed that each student when fires a bullet from the gun it is pushed back due to some force .

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Initial velocity , u = Vr = -2.5 m/s

Final velocity , v = 0 m/s

Finally goes to rest

Acceleration , a = ? m/s²

Displacement , s = - 5 m

Since moves through negative direction

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (0)² - (-2.5)² = 2a(-5)

⇒ 0 - 12.5 = -10a

⇒ 10a = 12.5

⇒ a = 1.25 m/s²

Apply formula for force ,

⇒ F = ma

⇒ F = (2)(1.25)

⇒ F = 2.5 N

Magnitude of the force on the gun is 2.5 N

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