a 35 kg bullet strikes a 5 kg wooden block and embeds itself in the block. the block and bullet slide off together at 8.6 m/s. what was the original velocity of the bullet?
Write the linear momentum conservation equation and solve for the unknown bullet velocity.
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To find the original velocity of the bullet, we can use the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.
Momentum is calculated by multiplying mass and velocity. Let's denote the original velocity of the bullet as v and the velocity of the block (and embedded bullet) after the collision as V.
Before the collision, the bullet is the only object moving, so its momentum is given by:
Momentum of the bullet before collision = mass of the bullet * original velocity of the bullet
P(before) = m(bullet) * v
Meanwhile, the block is initially at rest, so its momentum before the collision is zero:
P(before) = m(block) * 0 = 0
After the collision, both the block and bullet slide off together, so their combined momentum must be the total momentum after the collision:
P(after) = (mass of the block + mass of the bullet) * velocity of the block and bullet after collision
P(after) = (m(block) + m(bullet)) * V
Since momentum is conserved, we can equate the two expressions for momentum before and after the collision:
m(bullet) * v = (m(block) + m(bullet)) * V
Now we can plug in the given values:
m(bullet) = 35 kg (mass of the bullet)
m(block) = 5 kg (mass of the block)
V = 8.6 m/s (velocity of the block and bullet after collision)
35 kg * v = (5 kg + 35 kg) * 8.6 m/s
35v = 40 * 8.6 m/s
35v = 344 m/s
v ≈ 9.83 m/s
Therefore, the original velocity of the bullet was approximately 9.83 m/s.