A bullet with mass m_b = 10 g and initial velocity upsilon_0 strikes a stationar

Question

A bullet with mass m_b = 10 g and initial velocity upsilon_0 strikes a stationary wooden block with a mass M_B = 2.0 kg and gets embedded into the block. After the collision the bullet and block rise a vertical distance of 12 cm. There is no external forces during the collision. a) What is the initial momentum of the system? b) Obtain an expression for the final velocity of the bullet and block system after the collision. You answer should be in terms of m_b, M_B, and, upsilon_0. c) Write an expression for the kinetic energy of the system immediately after the collision?. You answer should be in terms of m_b, M_B, and, upsilon_0. d) What is the potential energy at the highest vertical displacement? e) What is the initial velocity of the bullet? What fraction of the initial kinetic energy of the bullet was lost in the collision (delta K/K_i)?

Explanation / Answer

Given that

mass of the bullet m1=0.010 kg

mass of the wooden block m2=2 kg

height h=0.12 m

(1)now we find the initial momentum

initial momentum=(m1+m2)(2gh)^1/2=(0.01+2)(2*9.8*0.12)^1/2=3.1 kg.m/s

(2)the finial velocity of bullet and woodenblock V=m1u1/m1+m2

(3)the kinetic energy after collision =(m1u1)^2/2(m1+m2)

(4) the initial velocity of the bullet u1=3.1/0.01=310 m/s

(5) fraction K'/K1=M2/(m1+m2) =2/2+0.01=0.995

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