A bullet of mass m travelling at speed vo in the direction shohwn above strikes a block of mass M and embeds itself in it. The block is sitting on the edge of a frictionless table of height H and is knocked off of the table by the collision.
a) What is the speed vb of the block immediately after the bullet sticks?
b) What distance R from the base of the table does the block land?
How long does it take to fall height H? figure that.
then,
distance=velocity*timefalling
solve for velocity
To find the answers to these questions, we can use the principles of conservation of momentum and conservation of mechanical energy.
a) What is the speed vb of the block immediately after the bullet sticks?
We can start by applying the principle of conservation of momentum. Since there is no external force acting horizontally and no friction, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision:
Initial momentum of the bullet (pb_initial) = m * vo
Initial momentum of the block (pb_initial) = 0 (since the block is at rest initially)
After the collision:
Final momentum of the bullet (pb_final) = m * vb (let vb be the speed of the block after the collision)
Final momentum of the block (pb_final) = M * vb
Conservation of momentum equation:
pb_initial = pb_final
m * vo = m * vb + M * vb
Simplifying the equation:
m * vo = (m + M) * vb
Solving for vb:
vb = (m * vo) / (m + M)
So, the speed of the block immediately after the bullet sticks is vb = (m * vo) / (m + M).
b) What distance R from the base of the table does the block land?
To find the distance R, we can use the principle of conservation of mechanical energy. The total mechanical energy before and after the collision must be equal.
Before the collision, the block has gravitational potential energy due to its height H: PE_initial = M * g * H
After the collision, when the block lands, it has no gravitational potential energy, but it gains kinetic energy: KE_final = (1/2) * M * vb^2
Conservation of mechanical energy equation:
PE_initial = KE_final
M * g * H = (1/2) * M * vb^2
Solving for R:
R = (vb^2 * H) / (2 * g)
So, the distance R from the base of the table where the block lands is R = (vb^2 * H) / (2 * g), where vb is the speed of the block immediately after the bullet sticks.