Calculating Block Velocity and Bullet Speed in Collision | Physics Homework

In summary: The block has a mass of 8.52g and the bullet has a mass of .458g. So the momentum of the system before the collision is 8.52*(.458+4.69) = 3.92m/s. The system's total momentum after the collision is 8.52+3.92=11.68m/s.
  • #1
Knfoster
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0

Homework Statement



An 8.52g bullet is fired into a 385g block that is initially at rest at the edge of a table of h = 1.03 m height. The bullet remains in the block, and after the impact the block lands d = 2.15 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table? What is the initial horizontal velocity of the block as it flies off the table? (assume this to be in the positive direction)

Determine the initial speed of the bullet.


Homework Equations


m1v1=m2v2


The Attempt at a Solution



The velocity of the red card (the one moving) is 0 m/s, and the velocity of e blue cart just after collision is 3.13 m/s. But I don't know how to go about determining the initial speed of the bullet.
 
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  • #2
Knfoster said:

Homework Statement



An 8.52g bullet is fired into a 385g block that is initially at rest at the edge of a table of h = 1.03 m height. The bullet remains in the block, and after the impact the block lands d = 2.15 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table? What is the initial horizontal velocity of the block as it flies off the table? (assume this to be in the positive direction)

Determine the initial speed of the bullet.


Homework Equations


m1v1=m2v2


The Attempt at a Solution



The velocity of the red card (the one moving) is 0 m/s, and the velocity of e blue cart just after collision is 3.13 m/s. But I don't know how to go about determining the initial speed of the bullet.

You're answering for a different problem. But that aside, maybe start with the parts of the question you know how to answer?

How long does it take for the block to drop 1.3m?
 
  • #3
Knfoster said:

Homework Statement



An 8.52g bullet is fired into a 385g block that is initially at rest at the edge of a table of h = 1.03 m height. The bullet remains in the block, and after the impact the block lands d = 2.15 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table? What is the initial horizontal velocity of the block as it flies off the table? (assume this to be in the positive direction)

Determine the initial speed of the bullet.

Apply conservation of linear momentum. For finding out the time it takes to reach the ground, I would opt to go for a kinematic equation since the range is also given
 
  • #4
sry. I was copying and pasting... I meant to say that I've figured that the time it takes the block to reach the und is .458 s, and the initial horizontal velocity of the block is 4.69 m/s... What I don't know how to figure is the initial speed of the bullet...?
 
  • #5
Knfoster said:
sry. I was copying and pasting... I meant to say that I've figured that the time it takes the block to reach the und is .458 s, and the initial horizontal velocity of the block is 4.69 m/s... What I don't know how to figure is the initial speed of the bullet...?

Well you have a perfectly inelastic collision.

You know the masses.
You know the final combined velocity.
You know the initial speed of the block at rest.
So ...
 
  • #6
so do I take the equation m1v1=(m1+m2)v2 ? And if so... what would my v2 be?
 
  • #7
Knfoster said:
so do I take the equation m1v1=(m1+m2)v2 ? And if so... what would my v2 be?

Haven't you found that out? :wink:
 
  • #8
I had a similar prob on mastering physics the other day,

Since

You know the masses.
You know the final combined velocity.
You know the initial speed of the block at rest.

then the total momentum= (mass block + mass bullet)final velocity

So the momentum of the block/bullet system is conserved. Therefore, the momentum before the collision is the same as the momentum after the collision. Find a second expression for total momentum , this time expressed as the total momentum of the system before the collision.

find this, set them equal to each other and solve for V
 
  • #9
THanks! I've figured it out. :)
 

1. What happens when two blocks collide?

When two blocks collide, they will exert a force on each other. This force can cause the blocks to accelerate, change direction, or break apart depending on their masses and speeds.

2. How does the conservation of momentum apply to two blocks colliding?

The conservation of momentum states that the total momentum of a system remains constant unless acted upon by an external force. In the case of two blocks colliding, the total momentum of both blocks before the collision will be equal to the total momentum after the collision.

3. What is the difference between elastic and inelastic collisions?

In an elastic collision, both kinetic energy and momentum are conserved. This means that after the collision, the blocks will have the same speed and direction as before. In an inelastic collision, some kinetic energy is lost and converted into other forms of energy, such as heat or sound.

4. How do the masses of the blocks affect the outcome of the collision?

The masses of the blocks play a crucial role in determining the outcome of a collision. A heavier block will have a greater inertia and will be more difficult to accelerate or change direction. Therefore, the lighter block will experience a greater change in velocity compared to the heavier block.

5. Why do objects stick together after a perfectly inelastic collision?

In a perfectly inelastic collision, the two objects stick together after the collision and move as one. This happens because the kinetic energy lost during the collision is converted into other forms of energy, such as deformation of the objects or sound. This energy is not enough to separate the objects, causing them to stick together.

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