Question + Problem on Collision/momentum

  • Thread starter rawimpact
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In summary: So V2'=5.7m/sWe then use this velocity as the initial because the block starts moving up the ramp. It will stop on the ramp, so final velocity V' = 0 m/s. We have to find the acceleration, and to make this easier we can make the problem one demensional. We set the ramp's surface as the X-axis and then find find Wx (force due to gravity pushing the block down the ramp). The weight = (1.3+0.05)(9.8)= 13.23. I set the problem so moving up the ramp is in the positive direction. so the summation problem looks like -Wx=M*A. I
  • #1
rawimpact
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1. A block of mass 1.3kg is resting at the base of a frictionless 40-degree ramp. A bullet of mass 50g is traveling parallel to the ramp's surface at 250 m/s. It collides with the block, enters it, and exits the other side at 100 m/s. How far up the ramp will the block travel?


http://img26.imageshack.us/img26/7883/60347116.png


2. M1V1+M2V2=M1V1'+M2V2' to solve for velocity. sumFx=M*A to find acceleration, and V'^2=V^2 +2adX, V = initial velocity, V' = final velocity, dX = delta or change in distance, and a = acceleration



3. So first we need to find the velocity of the block, so we use the first equation (M1V1+M2V2=M1V1'+M2V2' ) to get: (0.05)(250)+(1.3)(0) = (0.05)(100)+(1.3)(V2'), solving it gives us V2'=5.7m/s

We then use this velocity as the initial because the block starts moving up the ramp. It will stop on the ramp, so final velocity V' = 0 m/s. We have to find the acceleration, and to make this easier we can make the problem one demensional. We set the ramp's surface as the X-axis and then find find Wx (force due to gravity pushing the block down the ramp). The weight = (1.3+0.05)(9.8)= 13.23. I set the problem so moving up the ramp is in the positive direction. so the summation problem looks like -Wx=M*A. I can now solve for acceleration as (-13.23(sin(40)))/1.35 = a = -6.29. We have a negative acceleration because gravity is pushing the opposite direction. So now that i have V, V', and a, i can use the equation V'^2=V^2+2adX to get 0=5.7^2+2(-6.29)dX dX=2.58m


The answer key shows the answer to be 2.65, so is this due to rounding or a problem during my steps?
 
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  • #2
It's roundoff error--your method is fine.
rawimpact said:
3. So first we need to find the velocity of the block, so we use the first equation (M1V1+M2V2=M1V1'+M2V2' ) to get: (0.05)(250)+(1.3)(0) = (0.05)(100)+(1.3)(V2'), solving it gives us V2'=5.7m/s
Recalculate that speed. (Don't round off until the very last step in your calculations.)
 
  • #3
Thanks Doc
 

FAQ: Question + Problem on Collision/momentum

What is collision in terms of physics?

Collision in physics refers to the event where two or more objects come into contact with each other and exchange energy or momentum. This can occur between any type of matter, such as solid objects, fluids, or even subatomic particles.

How does momentum play a role in collisions?

Momentum is a measure of an object's motion and is defined as the product of its mass and velocity. In collisions, momentum is conserved, meaning that the total momentum of the objects before and after the collision remains the same. This concept is known as the law of conservation of momentum and is a fundamental principle in physics.

What is the difference between elastic and inelastic collisions?

In an elastic collision, both kinetic energy and momentum are conserved. This means that the objects involved bounce off each other and there is no loss of energy. In an inelastic collision, some of the kinetic energy is converted into other forms of energy, such as heat or sound, and is not conserved. However, momentum is still conserved in both types of collisions.

What factors affect the outcome of a collision?

The outcome of a collision is influenced by various factors, such as the masses and velocities of the objects involved, the angle of collision, and the elasticity of the objects. The type of collision, whether it is elastic or inelastic, also plays a significant role in determining the final outcome.

How is the concept of collision and momentum used in real-world applications?

The principles of collision and momentum are applied in various real-world scenarios, such as car crashes, sports, and space exploration. By understanding these concepts, scientists and engineers can design safer cars, improve athletic performance, and calculate the trajectory of spacecraft and satellites. They are also crucial in understanding the behavior of particles in particle accelerators and nuclear reactions.

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