A bullet with momentum of $10 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$ embeds itself into a large, heavy block of wood: $m _ { \text {wood } } = 12.6 \mathrm {~kg}$ and $m _ { \text {wood } } m _ { \text {bullet } }$. The block of wood is tied to a string connected to the ceiling, forming a ballistic pendulum. The initial velocity of the bullet was $1000 \mathrm {~m} / \mathrm { s }$. During the collision, the loss of mechanical energy is
A) 0 J.
B) 50 J.
C) 500 J.
D) 5000 J.

Question

A bullet with momentum of $10 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$ embeds itself into a large, heavy block of wood: $m _ { \text {wood } } = 12.6 \mathrm {~kg}$ and $m _ { \text {wood } } > > m _ { \text {bullet } }$. The block of wood is tied to a string connected to the ceiling, forming a ballistic pendulum. The initial velocity of the bullet was $1000 \mathrm {~m} / \mathrm { s }$. During the collision, the loss of mechanical energy is
A) 0 J.
B) 50 J.
C) 500 J.
D) 5000 J.

Quizplus · Accepted Answer

The loss of mechanical energy during the collision can be calculated by finding the difference between the initial kinetic energy of the bullet and the kinetic energy of the bullet-wood system immediately after the collision. The initial kinetic energy of the bullet is given by $\frac{1}{2}mv^2$, where $m$ is the mass of the bullet and $v$ is its velocity. Since the momentum of the bullet is given as $10 \, \text{kg} \cdot \text{m/s}$, and momentum $p = mv$, we can find the mass of the bullet as $m = \frac{p}{v} = \frac{10}{1000} = 0.01 \, \text{kg}$. The initial kinetic energy of the bullet is $\frac{1}{2} \times 0.01 \, \text{kg} \times (1000 \, \text{m/s})^2 = 5000 \, \text{J}$.After the collision, the bullet and wood move together with a common velocity $v'$, which can be found using conservation of momentum: $p_{\text{initial}} = p_{\text{final}}$, or $mv = (m_{\text{bullet}} + m_{\text{wood}})v'$. Since $v' = \frac{p_{\text{initial}}}{m_{\text{bullet}} + m_{\text{wood}}} = \frac{10}{0.01 + 12.6} \approx 0.793 \, \text{m/s}$, the kinetic energy of the bullet-wood system immediately after the collision is $\frac{1}{2}(0.01 + 12.6) \times (0.793)^2 \approx 5 \, \text{J}$.The loss of mechanical energy is the initial kinetic energy minus the kinetic energy after the collision: $5000 \, \text{J} - 5 \, \text{J} = 4995 \, \text{J}$, which is approximately $5000 \, \text{J}$.

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A Bullet with Momentum Of $10 \mathrm {~kg} \cdot \mathrm { m } / \mathrm { s }$