(Ballistic Pendulum) A bullet with mass m is fired into a block of wood with mas
ID: 1473610 • Letter: #
Question
(Ballistic Pendulum)
A bullet with mass m is fired into a block of wood with mass M, suspended like a pendulum, and makes a completely inelastic collision with it.
After the impact of the bullet, the block swings up to a maximum height h. Given the values of h = 2.00 cm = 0.0200 m, m = 4.00 g = 0.00400 kg, and M = 1.00 kg, (a) What is the (initial) velocity v_x of the bullet in m/s? (b) What is the velocity V_x of the bullet/block system right after impact? (c) What is the kinetic energy of the bullet right before impact? (d) What is the kinetic energy of the bullet/block system right after impact? Where did the initial energy of the bullet go? Please Note: Use Conservation of Linear Momentum to analyze the impact. In addition, use Conservation of Total Mechanical Energy to see how high the bullet/block system swings
Explanation / Answer
a)
(m+M)gh = 1/2mv2
v2 = 2*1.004*9.8*0.02/0.004
v = 9.919 m/s = v_x(initial)
b)
momentum conservation
m*v_x + M*0 = (m+M)*V_x
=> V_x = 0.004*9.919/1.004 = 0.0395 m/s
c)
the kinetic energy of the bullet right before impact = 1/2mv_x2 = 0.5*0.004*98.392 = 0.1968 J
d)
the kinetic energy of the bullet/block system right after impact = 1/2(m+M)*V_x2 = 0.5*1.004*0.0395 = 0.0198 J
energy lost in collision.
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