A pellet gun is fired at a cardboard box of mass m_2 = 0.75 kg on a frictionless

Question

A pellet gun is fired at a cardboard box of mass m_2 = 0.75 kg on a frictionless surface. The pellet has a mass of m_1 = 0.018 kg and flys at a velocity of v_1 = 63 m/s. It is observed that the box is moving at a velocity of v_2 = 0.11 m/s after the pellet passes through it. Write an expression for the magnitude of the pellet's velocity as it exits the box v_f. What is the pellet's final velocity v_f, in meters per second? If the pellet doesn't exit the box, what will the velocity of the box v_2^', be in meters per second?

Explanation / Answer

a)

Given that

m1 =0.018kg

m2 =0.75kg

u1 =63m/s

v2 =0.11m/s

Now from the conservation of momentum after the pellet and cardboard is

m1u1+m2u2 =m1v1+m2v2

Here the card board is atrest (u2) =0

Now from thecondition we get

m1u1=m1v1+m2v2

Therefore v1 =u1 -(m2/m1)v2

b)

Now the pellet's final velocity is given by

v1 =u1 -(m2/m1)v2

=63 -(0.75/0.018)*(0.11m/s)

=58.416m/s

c)

Ifthe pellet does'nt exit the box what will be the velocity of the box

Then the final velocity of pellet and cardboard will be common i.e v1=v2 =V

m1u1+mu2 =(m1+m2)V

here u2 =0

Now m1u1 =(m1+m2)V

V =v2' =m1u1/(m1+m2) =0.018*63/(0.018+0.75) =1.476m/s

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