A 29 g rubber bullet is travelling to the right at 200 m/s, when it hits a 1.1 kg block of wood sitting on a horizontal frictionless surface. If this collision is inelastic (though not perfectly inelastic), and the velocity of the block of wood after the collision is 6.85 m/s to the right, what is the velocity of the bullet after the collision? Give your answer in m/s and treat right as the positive direction and left as the negative direction. (Include the sign in your answer if it is a negative value, and no sign if the answer is positive.)
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i'm just unsure of what formula i should apply!
so i found a formula: m1v1ix +m2v2ix = m1v1fx + m2v2fx
which i rearranged to find v1fx and eliminated m2v2ix because m2v2ix = 0
so the eqn looked like: v1fx = (m1v1ix - m2v2fx)/m1
subbing in i got: [(0.029)(200) - (1.1)(6.25)]/(0.029)
which gave the answer -59.8276. is this correct?
You are using the right equation, but the 6.25 does not agree with the original 6.85 for v2fx
To find the velocity of the bullet after the collision, we need to use the principle of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.
The momentum of an object is defined as the product of its mass and velocity. So we can calculate the initial momentum (before the collision) using the bullet's mass and velocity:
Initial momentum of the bullet = mass of the bullet × velocity of the bullet
P_initial = (29 g) × (200 m/s)
We'll need to convert the mass of the bullet to kilograms since the other quantities are in SI units:
P_initial = (0.029 kg) × (200 m/s)
The momentum after the collision is the sum of the momenta of the bullet and the block:
P_after = (mass of the bullet × velocity of the bullet) + (mass of the block × velocity of the block)
Since it is an inelastic collision, the bullet and the block stick together after the collision, sharing a common final velocity. So we can replace the velocity of the block with the common velocity of the bullet and block:
P_after = (mass of the bullet + mass of the block) × (common velocity of bullet and block)
We can now set the initial and final momenta equal to each other to solve for the common velocity:
P_initial = P_after
(0.029 kg) × (200 m/s) = (0.029 kg + 1.1 kg) × (common velocity)
Let's solve for the common velocity:
(0.029 kg) × (200 m/s) = (1.129 kg) × (common velocity)
5.8 kg·m/s = 1.129 kg × (common velocity)
Now, let's solve for the common velocity:
common velocity = (5.8 kg·m/s) / (1.129 kg)