Question

An 0.025 kg bullet traveling at 700 m/s strikes and passes through the center of mass of a 0.50 kg block of wood that is initially at rest on a smooth flat surface. The bullet passes through the block and emerges from the other side traveling at 300 m/s? How fast will the block be sliding just after the bullet emerges, and how much energy (in Joules) will be converted to heat? (Neglect the effect of sliding friction during the impact.)

Answer 1

Given Mass of bullet.m_b = 0.025 kg

initial velocity of bullet , u_b = 700 m/s

Final Velocity of bullet ,v_b = 300m/s

Mass of wooden block.m_w = 0.025 kg

initial velocity of wooden block , u_w = 0 m/s

Final Velocity of wooden block ,v_w = ?

Here Momentum is conserved

so Initial momentum =Final momentum

m_b*u_b +m_w*u_w = m_b*v_b +m_w*v_w

0.025*700 + 0.5 *0 = 0.025*300 +0.5*v_w

v_w = 0.025*400/0.5

v_w = 20m/s

ANSWER: Velocity of wooden block after bullet emerges =20m/s

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Now lets find the energy converted into heat

Initial Kinetic energy = 1/2m_b*u_b² +1/2m_wu_w²

KE_i = 1/2 *0.025*700² +0

KE_i = 6125 J

Final Kinetic Energy = 1/2m_b*v_b² +1/2m_wv_w²

KE_f = 1/2*0.025*300² +1/2*0.5*20²

KE_f = 1225 J

Therefore energy converted into heat = KE_i - KE_f

= 6125 - 1225

=4900 J

ANSWER = 4900 J

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