A fast moving object of mass 200g travels at 100m/s and hits a block of wood of mass 2kg. The two bodies moved together after impact. Find the velocity with which they moved together after collision.?
momentum is conserved
.2 * 100 = (2 + .2) * v
To find the velocity with which the two bodies move together after the collision, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision, as long as no external forces act on the system.
Let's denote the initial velocity of the fast-moving object as v1 (given as 100m/s) and its mass as m1 (given as 200g, which is 0.2kg). The velocity of the block of wood before the collision is 0 since it is at rest, and its mass is m2 (given as 2kg).
Using the principle of conservation of momentum, we can write:
(m1 * v1) + (m2 * v2) = (m1 + m2) * vf
Where v2 is the velocity of the block of wood after the collision (which we need to find) and vf is the final velocity of the combined system.
Let's substitute the known values into the equation:
(0.2kg * 100m/s) + (2kg * v2) = (0.2kg + 2kg) * vf
20kg⋅m/s + 2kg⋅v2 = 2.2kg⋅vf
Now, we need to solve for v2. Rearranging the equation, we have:
2kg⋅v2 = 2.2kg⋅vf - 20kg⋅m/s
v2 = (2.2kg⋅vf - 20kg⋅m/s) / 2kg
Note that the kg⋅m/s terms cancel out, leaving us with:
v2 = (2.2vf - 20) / 2
Therefore, the velocity with which the two bodies move together after the collision is given by (2.2vf - 20) / 2.