A 16.0-g bullet is shot into a wooden block at rest with mass 4550 g on a frictionless surface. The bullet strikes and is lodged in the block, causing is to slide at a speed of 1.20 m/s. Find the speed of the bullet (in m/s) before striking the block.
conservation of momentum applies
.016*V=(.016+4.550)*1.2
solve for V
To find the speed of the bullet before striking the block, we can use the principle of conservation of momentum. The equation for conservation of momentum can be written as follows:
(m1 * v1) + (m2 * v2) = (m1 + m2) * vf
Where:
m1 = mass of the bullet = 16.0 g = 0.016 kg
v1 = velocity of the bullet before striking = ?
m2 = mass of the block = 4550 g = 4.550 kg
v2 = velocity of the block after the bullet is lodged = 1.20 m/s
vf = final velocity of the bullet and block together
Since the wooden block is at rest initially, its initial velocity is zero. The equation can be simplified to:
(m1 * v1) = (m1 + m2) * vf
Substituting the known values:
(0.016 kg * v1) = (0.016 kg + 4.550 kg) * 1.20 m/s
Solving for v1:
0.016 kg * v1 = 5.566 kg * 1.20 m/s
v1 = (5.566 kg * 1.20 m/s) / 0.016 kg
v1 = 417.9 m/s
Therefore, the speed of the bullet before striking the block is approximately 417.9 m/s.
To find the speed of the bullet before striking the block, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it.
In this case, the bullet and the block form an isolated system (ignoring any external forces like air resistance) before the collision. Therefore, the total momentum before the collision is equal to the total momentum after the collision.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v), so we can write the equation as:
Total momentum before collision = Total momentum after collision
(mass of the bullet x velocity of the bullet before collision) + (mass of the block x velocity of the block before collision) = (mass of the bullet + mass of the block) x (velocity of the block after collision)
Let's plug in the given values:
(16.0 g) x (v) + (4550 g) x (0 m/s) = (16.0 g + 4550 g) x (1.20 m/s)
First, we need to convert the masses from grams to kilograms, since the SI unit for mass is kilograms:
(0.016 kg) x (v) + (4.550 kg) x (0 m/s) = (0.016 kg + 4.550 kg) x (1.20 m/s)
Now, we can simplify the equation:
0.016v + 0 = (4.566 kg) x (1.20 m/s)
0.016v = 5.48
v = 5.48 / 0.016
v = 342.5 m/s
Therefore, the speed of the bullet before striking the block is approximately 342.5 m/s.