An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height (Fig. P6.26). The bullet remains in the block, and after the impact the block lands 2.00 m from the bottom of the table. Determine the initial speed of the bullet.

So the block fell through a height of 1 m in time t, where

1.0=(1/2)gt²
or
t=√(2/9.81)
=0.452 s

During this time, the block has travelled horizontally 2m, or
horiz. velocity
=2/0.452
=4.43 m/s

This velocity, v, is the velocity of the block after impact, given by the equation of momentum before and after impact, namely
m1u1+m2u2=(m1+m2)v
or
v=(m1u1+m2u2)/(m1+m2)
where
m1=8g
m2=250g
u1= to be determined
u2=0 (block)
Solve for u1 (velocity of bullet)

do the end first

mass m (does not matter what m is) lands 2 meters from the table

how long did it fall?
h = 0 at end
0 = 1 - (1/2)(9.8) t^2
2 = 9.8 t^2
t = .45 seconds to fall to floor off table = time in air
distance = speed*time
2 = speed* .45
speed = 4.4 m/s horizontal
.008 v = .258 (4.4)
v = 141.9 m/s

You have to use kinematics then you can use the collision equation. The kinematics equation you use is vfysq=viySq+2a🔻y

Collison equation is my+mv=mv+mv

To determine the initial speed of the bullet, we need to use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

Let's break down the problem step by step:

Step 1: Calculate the momentum of the block before the collision.
The momentum of an object is defined as the product of its mass and velocity: momentum = mass × velocity. Since the block is initially at rest, its momentum before the collision is zero.

Step 2: Calculate the momentum of the bullet before the collision.
The momentum of the bullet before the collision can be calculated using the same formula: momentum = mass × velocity. The mass of the bullet is given as 8.00 g, which is equivalent to 0.008 kg. We need to find the velocity of the bullet before the collision.

Step 3: Calculate the momentum of the block and the bullet after the collision.
After the collision, the bullet remains embedded in the block, so they move together as one system. To calculate their combined momentum after the collision, we need to find the velocity of the block and bullet together.

Step 4: Use the principle of conservation of momentum.
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. So, we can set up an equation:

(total momentum before collision) = (total momentum after collision)

Since the momentum of the block before the collision is zero, the equation becomes:
(momentum of the bullet before the collision) = (momentum of the block and bullet after the collision)

Step 5: Solve for the initial velocity of the bullet.
Rearrange the equation to solve for the initial velocity of the bullet before the collision.

Let's go ahead and calculate the initial speed of the bullet using the information provided.

Step 1: Calculate the momentum of the block before the collision.
The momentum of the block before the collision is zero since it is initially at rest.

Step 2: Calculate the momentum of the bullet before the collision.
mass of the bullet = 8.00 g = 0.008 kg (converting grams to kilograms)
velocity of the bullet = ?
momentum of the bullet before the collision = mass of the bullet × velocity of the bullet

Step 3: Calculate the momentum of the block and the bullet after the collision.
mass of the block and bullet after the collision = mass of the block + mass of the bullet
velocity of the block and bullet after the collision = ?
momentum of the block and bullet after the collision = (mass of the block and bullet after the collision) × (velocity of the block and bullet after the collision)

Step 4: Use the principle of conservation of momentum.
(momentum of the bullet before the collision) = (momentum of the block and bullet after the collision)

Step 5: Solve for the initial velocity of the bullet.
Rearrange the equation to solve for the initial velocity of the bullet before the collision.

I want answer for the above question

An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height (Fig. P6.26). The bullet remains in the block, and after the i

48.Consider a frictionless track as shown in Figure. A block of mass m1 = 5.00 kg is released from A. It makes a head-on elastic collision at B with a block of mass m2 =10.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.