Assignment 20 Query

Your work on this assignment is good.

Let me know if anything is unclear, and include specifics about what you do and do not understand.

......!!!!!!!!...................................

22:54:57 Explain how we get the components of the resultant of two vectors from the components of the original vectors.

......!!!!!!!!...................................

RESPONSE --> by using the formula Magnitude * sin(angle) for y Magnitude * cos(angle) for x

.................................................

......!!!!!!!!...................................

22:55:54 ** If we add the x components of the original two vectors we get the x component of the resultant. If we add the y components of the original two vectors we get the y component of the resultant. **

......!!!!!!!!...................................

RESPONSE --> I guess I missed the question...but to add on to my previous response you can add the resulting x components together for the value and then add the resulting y components together.

.................................................

......!!!!!!!!...................................

22:57:01 Explain how we get the components of a vector from its angle and magnitude.

......!!!!!!!!...................................

RESPONSE --> This goes back to my reponse in the previous question which would apply here magnitude * cos(angle) for x component magnitude * sin(anlge) for the y component

.................................................

......!!!!!!!!...................................

22:57:07 ** To get the y component we multiply the magnitude by the sine of the angle of the vector (as measured counterclockwise from the positive x axis). To get the x component we multiply the magnitude by the cosine of the angle of the vector (as measured counterclockwise from the positive x axis). **

......!!!!!!!!...................................

RESPONSE --> ok

.................................................

......!!!!!!!!...................................

23:41:17 gen phy #7.12 23 g bullet 230 m/s 2-kg block emerges at 170 m/s speed of block

......!!!!!!!!...................................

RESPONSE --> I struggled with this problem a bit but using the book as a reference....I ended up using the momentum before = momentum after equation and obtain b is the bullet and c is the block Pi = Mb*vb Pf = (Mb + mc)(vb)

This would be the momentum if both bullet and block were moving at velocity vb. However they are not moving at the same velocity either before or after collision, and the after-collision velocity of the bullet should be respresented as vb'.

In an inelastic collision the bullet and block have the same velocity after collision. In this collision their velocities are different both before and after collision.

So your solution is pretty much following the scheme for inelastic collisions, where Pf = (m1 + m2) v ', but this is not an inelastic collision.

.................................................

......!!!!!!!!...................................

23:45:52 **STUDENT SOLUTION: Momentum conservation gives us m1 v1 + m2 v2 = m1 v1' + m2 v2' so if we let v2' = v, we have (.023kg)(230m/s)+(2kg)(0m/s) = (.023kg)(170m/s)+(2kg)(v). Solving for v: (5.29kg m/s)+0 = (3.91 kg m/s)+(2kg)(v) .78kg m/s = 2kg * v v = 1.38 kg m/s / (2 kg) = .69 m/s. INSTRUCTOR COMMENT: It's probably easier to solve for the variable v2 ': Starting with m1 v1 + m2 v2 = m1 v1 ' + m2 v2 ' we add -m1 v1 ' to both sides to get m1 v1 + m2 v2 - m1 v1 ' = m2 v2 ', then multiply both sides by 1 / m2 to get v2 ` = (m1 v1 + m2 v2 - m1 v1 ' ) / m2. Substituting for m1, v1, m2, v2 we will get the result you obtained.**

......!!!!!!!!...................................

RESPONSE --> I understand this problem much more clearly...I seem to follow your explinations and details well..but the book seem to confuse me a little bit as it seems difficult to follow....after your explination I have a good grasp on the problem ...and will actully rework it a few times with different variables to make sure I get the hang of it..

.................................................

......!!!!!!!!...................................

23:45:58 **** Univ. 8.70 (8.68 10th edition). 8 g bullet into .992 kg block, compresses spring 15 cm. .75 N force compresses .25 cm. Vel of block just after impact, vel of bullet?

......!!!!!!!!...................................

RESPONSE --> Not Required

.................................................

......!!!!!!!!...................................

23:46:01 ** The spring ideally obeys Hook's Law F = k x. It follows that k = .75 N / .25 cm = 3 N / cm. At compression 15 cm the potential energy of the system is PE = .5 k x^2 = .5 * 3 N/cm * (15 cm)^2 = 337.5 N cm, or 3.375 N m = 3.375 Joules, which we round to three significant figures to get 3.38 J. The KE of the 1 kg mass (block + bullet) just after impact is in the ideal case all converted to this PE so the velocity of the block was v such that .5 m v^2 = PE, or v = sqrt(2 PE / m) = sqrt(2 * 3.38 J / ( 1 kg) ) = 2.6 m/s, approx. The momentum of the 1 kg mass was therefore 2.6 m/s * .992 kg = 2.6 kg m/s, approx., just after collision with the bullet. Just before collision the momentum of the block was zero so by conservation of momentum the momentum of the bullet was 2.6 kg m/s. So we have mBullet * vBullet = 2.6 kg m/s and vBullet = 2.6 kg m/s / (.008 kg) = 330 m/s, approx. **

......!!!!!!!!...................................

RESPONSE --> ok

.................................................

"