Impulse and Momentum Problem

In summary: Bx = 400 cos 20 By = 400 sin 20 Cx = ? Cy = ? In summary, using the given information and equations for conservation of momentum, the velocity of the third piece, C, can be calculated by resolving the momentum vectors of pieces A and B. Based on the diagram, the values for A and B should be 500 cos 30 and 500 sin 30 for the x and y components, respectively, and 400 cos 20 and 400 sin 20 for the x and y components of B. The final velocity for C can then be calculated using the equation v3 = p3/m3.
  • #1
bgaspar00
1
0
A grenade of mass 10 kg explodes into three pieces in the same plane, two of which, A (5.0 kg) and B (2.0 kg), move off as shown. Calculate the velocity of the 3.0 kg third piece, C.

I cannot show all of my work becuz it is basically just a sketch of what the diagram would look like, then drawing vectors and so on. However, i cannot get the right answer.
So here's the diagram it shows...
http://www.cbv.ns.ca/rv/physics/Physics12/BLM10-3.pdf


Δp = 0 where pi = 0

pix = pfx and piy = pfy






p1’



p1y’ = Cos30(5.0 kg (100.0 m/s))

= 433.013 kgm/s

p1x’ = Sin30(5.0(100m/s))
= 250


p2y’ = cos20 (2kg)(200 m/s)

= cos20 ( 400 kgm/s)

p2x’ = sin20 (400 kgm/s)




p1x’ + p2x’ + p3x’ = 0

500 + (- sin 20 (400 kgm/s)) + p3x’ = 0

p3x’ = 363.192 kgm/s

p1y’ + p2y’ + p3y’ = 0

500 kgm/s + (- cos 20(400 kgm/s)) + p3y’ = 0

p3y’ = 124.123kgm/s
p3 is found by resolving p3x’ and p3y’



p3x’

p3y’

p3

θ


p3^2 = [363.192kgm/s]2

+[124.123 kgm/s]2

|p3| = 383.816 kgm/s

θ = tan-1 [(124.123kgm/s)/(363.192kgm/s)]

= 19 degrees? wtf v3 = p3/m3

v3 = 383.816kgm/s/(3kg)

v3 = 128 m/s

This is completely wrong the answer is suppose to be V3= 128.92 m/s (S 82 E) or something...soo where did i go wrong? and is there an easier wayyyyy!??
 
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  • #2
I'm denied permission to view the file.
 
  • #3
I think you've mixed up your sines and cosines. The 'x' components should be cosine values, and the 'y' components should be sine values. To start you off, based on the picture:

Ax = - (500 cos 30)
Ay = 500 sin 30
 

1. What is impulse and momentum?

Impulse and momentum are two related concepts in physics. Impulse is defined as the force applied to an object over a certain period of time, while momentum is the measure of an object's motion and is calculated by multiplying its mass by its velocity.

2. How are impulse and momentum related?

Impulse and momentum are related through Newton's second law of motion, which states that the change in an object's momentum is equal to the impulse applied to it. This means that the larger the impulse applied to an object, the greater its change in momentum will be.

3. What is the equation for calculating impulse?

The equation for calculating impulse is impulse = force x time, or J = FΔt, where J is impulse, F is force, and Δt is the time interval over which the force is applied. Impulse is measured in units of Newton-seconds (N-s).

4. How is momentum conserved in a system?

Momentum is conserved in a system when the total momentum of all the objects in the system remains constant. This means that the initial momentum of the system must equal the final momentum of the system, even if individual objects within the system may experience changes in momentum.

5. What are some real-world applications of impulse and momentum?

Impulse and momentum are important concepts in many real-world scenarios, such as car crashes, sports, and rocket propulsion. In car crashes, the impulse of the impact can affect the momentum of the vehicles involved. In sports, athletes use their momentum to increase the force of their movements. And in rocket propulsion, the impulse of the exhaust gases creates a reaction force that propels the rocket forward.

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