Answer to Question #91208 in Physics for TFercado

Question

A bullet of mass m=0.0030 kg collides with and becomes embedded in a block of mass M = 1.5 kg. The block and bullet rise to a height h= 0.40 m as shown. Note that the momentum of the bullet/block combination immediately after the collision must be the same as the momentum of just the bullet immediately before the collision.
a.Use energy considerations to determine the velocity of the block immediately after the
collision.
b.Determine the velocity of the bullet just before the collision.
c.What percentage of the initial kinetic energy is lost during the collision?
d.If the bullet were to pass through the block instead of becoming embedded, would the block rise to a height higher than h, lower than h or equal to h? Explain your thinking in a sentence or two (no more).

Answer 1

a. From the conservation of energy:

$(m+M)gh=0.5(m+M)V^2$

$V=\sqrt{(2gh)}=\sqrt{(2(9.8)(0.4))}=2.8\frac{m}{s}$

b. From the conservation of momentum:

$mv=(m+M)V$

$v=(1+\frac{M}{m})V=\left(1+\frac{1.5}{0.003}\right)2.8=1400\frac{m}{s}$

c.

$E_{lost}=0.5mv^2-0.5(m+M)V^2$

$\frac{E_{lost}}{E}=\frac{0.5(0.003)(1400)^2-0.5(0.003+1.5)2.8^2}{0.5(0.003)(1400)^2}=0.998=99.8\%$

d. It rise to a height lower than h. It is because some energy will be the kinetic energy of the bullet.