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Calculate the recoil velocity in the horizontal direction, in meters per second, of a 1.25-kg plunger that directly interacts with a 0.0175-kg bullet fired at 585 m/s from the gun. Take the firing direction to be the positive direction.

Respuesta :

Answer:

[tex]v_1=-8.19\ m/s[/tex]'

Explanation:

It is given that,

Mass of the plunger, [tex]m_1=1.25\ kg[/tex]

Mass of the bullet, [tex]m_2=0.0175\ kg[/tex]

Initially both plunger and the bullet are at rest, [tex]u_1=u_2=0[/tex]

Final speed of the bullet, [tex]v_2=585\ m/s[/tex]

Let [tex]v_1[/tex] is the final speed of the plunger. Using the conservation of momentum to find it. The equation is as follows :

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

Since, [tex]u_1=u_2=0[/tex]

[tex]m_1v_1+m_2v_2=0[/tex]

[tex]v_1=-\dfrac{m_2v_2}{m_1}[/tex]

[tex]v_1=-\dfrac{0.0175\times 585}{1.25}[/tex]

[tex]v_1=-8.19\ m/s[/tex]

So, the recoil velocity of the plunger is 8.19 m/s. Hence, this is the required solution.

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