Change in speed of the bullet after striking an object

In summary: Combining these two equations (from momentum and kinetic energy conservation), we get: \mu_1+MV_1=mu_2+MV_2+KV_2\ where K is the kinetic energy of the bullet.
  • #1
omc1
100
0

Homework Statement


A gangster fires a bullet (m1 = 30 g, v1 = 190 m/s) at Stupendous Man (m2 = 31 kg), but it simply bounces away elastically. If Stupendous man was standing on a frictionless surface, what is the change in speed of the bullet after striking stupendous man? Be careful with signs and units!


Homework Equations

conservation of momentum, mv1+mv2=mv1+mv2



The Attempt at a Solution

the man is not moving initially, and final, i got 0.183 for the final bullet speed but that's not right, Iam confused about how to find the final for the bullet?
 
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  • #2
omc1 said:

Homework Statement


A gangster fires a bullet (m1 = 30 g, v1 = 190 m/s) at Stupendous Man (m2 = 31 kg), but it simply bounces away elastically. If Stupendous man was standing on a frictionless surface, what is the change in speed of the bullet after striking stupendous man? Be careful with signs and units!

Homework Equations

conservation of momentum, mv1+mv2=mv1+mv2

The Attempt at a Solution

the man is not moving initially, and final, i got 0.183 for the final bullet speed but that's not right, I am confused about how to find the final for the bullet?
You haven't explained how you got that result, so it's hard to say where you went wrong.

The problem doesn't state the direction in which the bullet bounces away, so I would assume that the bullet goes back toward the location it was fired from.

Did you use the fact that it's an elastic collision ?

Such a problem can often be solved most easily in the center of mass reference frame. If you do, don't round-off anything until your final answer.
 
  • #3
i thought i was using the formula for elastic collision, .03*190=.03*V2+31*V2, but I though the man did not move so I am confused about that, how would I apply the cm to this?
 
  • #4
omc1 said:
i thought i was using the formula for elastic collision, .03*190=.03*V2+31*V2, but I though the man did not move so I am confused about that, how would I apply the cm to this?
That is for an inelastic collision. That's evident because you have both the bullet and the man traveling at the same velocity.
 
  • #5
but i don't know either final velocity...
 
  • #6
Is the answer 0.367ms-1.

Soln:
We have two equations (one from momentum conservation and the other from kinetic
energy conservation) with two variables (final velocities of the bullet and the man).Solve these
two equations and get the final velocity of the man in terms of initial velocity of both and their masses.

The final equation will be:

V=M2U2/(M1+M2)

where M1 is the mass of the man,
M2 is the mass of the bullet,
U2 is the velocity of the bullet.
 
  • #7
yes, that is the answer, why is the mass combined because the bullet bounces off.
 
  • #8
omc1 said:

Homework Statement


A gangster fires a bullet ...

Homework Equations

conservation of momentum, mv1+mv2=mv1+mv2
...

In your above equation (from your Original Post), you have two unknowns. You need another equation in order to get a solution.

Let's call the final velocity of the man, V2, and the final velocity of the bullet, u2.

The initial velocities being V1 = 0 m/s and u1 = 190 m/s .

Similarly, let M be the mass of the man, M = 31 kg. Let m be the mass of the bullet, m = 30 grams = 0.03 kg.

(I took it upon myself to change some variable names.)

Conservation of momentum gives [itex]\ mu_1+MV_1=mu_2+MV_2\ [/itex] where V1=0, for the (initially) stationary man.
[itex]\ mu_1=mu_2+MV_2\ [/itex]​

The other equation you need is from conservation of kinetic energy (It's an elastic collision.)
[itex]\ (1/2)m{u_1}^2=(1/2)m{u_2}^2+(1/2)M{V_2}^2\ [/itex]​
 

1. How does the weight of the bullet affect its speed after striking an object?

The weight of the bullet does not affect its speed after striking an object. The speed of the bullet is primarily determined by its initial velocity and the resistance it encounters upon impact.

2. Does the material of the object being struck affect the change in speed of the bullet?

Yes, the material of the object being struck can affect the change in speed of the bullet. Objects with a higher density or hardness will cause more resistance and result in a greater decrease in the bullet's speed.

3. Is the angle at which the bullet strikes the object important in determining its change in speed?

Yes, the angle at which the bullet strikes the object can affect its change in speed. A direct, perpendicular impact will result in a greater decrease in speed compared to a glancing blow at an angle.

4. What role does air resistance play in the change in speed of the bullet after striking an object?

Air resistance can significantly affect the change in speed of a bullet after striking an object, especially at high velocities. The drag force of the air can slow down the bullet and cause a greater decrease in speed upon impact.

5. Can the shape or design of the bullet impact its change in speed after striking an object?

Yes, the shape and design of the bullet can affect its change in speed after striking an object. A more streamlined bullet will encounter less resistance and maintain its speed better upon impact, while a less aerodynamic design may result in a greater decrease in speed.

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