Based on conservation of linear momentum, if m1 = m2, V1 = 2.8 m/s, V2 =...
Based on conservation of linear momentum, if m1 = m2, V1 = 2.8 m/s, V2 = 2.4 m/s, and LaTeX: \theta ? 1 = 22 degrees, what will be the angle LaTeX: \theta ? 2 (in degrees)?
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Based on conservation of linear momentum, if m1 = m2, V1 = 2.8
m/s, V2 =...
conservation of momentum before v1=2 m1=100g v2=-3 m2= 300 find after u1=-2 u2=___?
conservation of momentum before v1=2 m1=100g v2=-3 m2= 300 find after u1=-2 u2=___?
m1 m2 V1 v2 Ptotal V2 1.8 m/s 1.2 kg 1.2 kg +1.5 m/s 1.8 m/s...
m1 m2 V1 v2 Ptotal V2 1.8 m/s 1.2 kg 1.2 kg +1.5 m/s 1.8 m/s 2.4 kg 4.8 kg +1.3 m/s 0m/s 2.5 kg 3.9 kg 5.1 kg 1 kg 0.9 m/s 4.6 m/s -0.433 m/s 0.85 m/s 11.5 2.75 m/s kg m/s 4.6 m/s Table 1
V1 = 8 m/s V2 = 4 m/s Vi' = ? v2 = 7 m/s OW...
V1 = 8 m/s V2 = 4 m/s Vi' = ? v2 = 7 m/s OW m1 m2 mi m2 In the above collision what is va'? Question 7 8 pt [Continued from Question 6.] If my = 350 g what is m2?
Two masses collide elastically (hit & bounce) where m1 = 0.5 kg, m2 = 1.5 kg,...
Two masses collide elastically (hit & bounce) where m1 = 0.5 kg, m2 = 1.5 kg, v1 = 1 m/s, v2 = 0 m/s Calculate the speeds of the balls after the collisions by using the formulas for elastic collisions: v1' = [v1 * (m1-m2) / (m1+m2)] + [v2 * (2m2) / (m1+m2)] v2' = [v1 * (2m1) / (m1+m2)] - [v2 * (m1-m2) / (m1+m2)]
The show system is orbiting counterclockwise around stationary point O, r=2.4 m m1=1.5kg, m2=1.8kg, m3=2.3kg Linear...
The show system is orbiting counterclockwise around stationary
point O, r=2.4 m
m1=1.5kg, m2=1.8kg, m3=2.3kg
Linear velocity is the same for each particle and it is 2.8m/s at
the presented point of time
Theta is 12 degrees
overall velocity of the entire system as a single vector =?
overall acceleration of the entire system as a single
vector?=?
total kinetic energy of this system=?
mi ma м»
Two objects with masses represented by m1 and m2 are moving such that their combined total...
Two objects with masses represented by m1 and m2 are moving such that their combined total momentum has a magnitude of 18.5 kg · m/s and points in a direction 71.5°above the positive x-axis. Object m1 is moving in the x direction with a speed of v1 = 2.75 m/s and m2 is moving in the y direction with a speed of v2 = 3.22 m/s. Determine the mass of each object in kilograms. m1= kg m2= kg
Problem C1: (Conservation of Momentum) Part C (LINEAR MOMENTUM A puck of mass m, = 3kg...
Problem C1: (Conservation of Momentum) Part C (LINEAR MOMENTUM A puck of mass m, = 3kg has an initial velocity of 10 m/s at 30 m2 = 5kg has a velocity of 5m/s at 45° W of N. They collide Om/s at 30° S of E. A second puck of mass They collide and stick together. V 30° Find the magnitude and direction after collision.
V1 = 8 m/s V2 = 4 m/s Vi' = ? V2 = 7 m/s mi...
V1 = 8 m/s V2 = 4 m/s Vi' = ? V2 = 7 m/s mi m2 Car ma m2 In the above collision what is vi'? 0 -8 m/s O 3 m/s O-1 m/s O 5 m/s O 1 m/s O-3 m/s O 8 m/s 0-2 m/ O -5 m/s O 4 m/s 0-4 m/s 0 2 m/s
A block of mass m1 = 1.10 kg moving at v1 = 1.20 m/s undergoes a...
A block of mass m1 = 1.10 kg moving at v1 = 1.20 m/s undergoes a completely inelastic collision with a stationary block of mass m2 = 0.900 kg . The blocks then move, stuck together, at speed v2. After a short time, the two-block system collides inelastically with a third block, of mass m3 = 2.40 kg , which is initially at rest. The three blocks then move, stuck together, with speed v3. Assume that the blocks slide without...
Solve for v1 and v2 given conservation of kinetic energy Search this book n, in this...
Solve for v1 and v2 given conservation of kinetic
energy
Search this book n, in this case, reads Pi = Pf | M2V2,i = mjº18+ m2V2,f. ergy reads K; = K mavzi = {m101s+ mauz.s. lons in two unknowns. The algebra is tedious but not terribly difficult; you de (mi-m2)+1,i+2m2 U2, mi + m2 (m2-mi U2,+2m, V1, mi +m2 U2f numbers, we obtain V1,8 V2,1 = = 2.22 m -0.28 m
m1 m2 V1 v2 Ptotal V2 1.8 m/s 1.2 kg 1.2 kg +1.5 m/s 1.8 m/s 2.4 kg 4.8 kg +1.3 m/s 0m/s 2.5 kg 3.9 kg 5.1 kg 1 kg 0.9 m/s 4.6 m/s -0.433 m/s 0.85 m/s 11.5 2.75 m/s kg m/s 4.6 m/s Table 1
V1 = 8 m/s V2 = 4 m/s Vi' = ? v2 = 7 m/s OW m1 m2 mi m2 In the above collision what is va'? Question 7 8 pt [Continued from Question 6.] If my = 350 g what is m2?
The show system is orbiting counterclockwise around stationary
point O, r=2.4 m
m1=1.5kg, m2=1.8kg, m3=2.3kg
Linear velocity is the same for each particle and it is 2.8m/s at
the presented point of time
Theta is 12 degrees
overall velocity of the entire system as a single vector =?
overall acceleration of the entire system as a single
vector?=?
total kinetic energy of this system=?
mi ma м»
Problem C1: (Conservation of Momentum) Part C (LINEAR MOMENTUM A puck of mass m, = 3kg has an initial velocity of 10 m/s at 30 m2 = 5kg has a velocity of 5m/s at 45° W of N. They collide Om/s at 30° S of E. A second puck of mass They collide and stick together. V 30° Find the magnitude and direction after collision.
V1 = 8 m/s V2 = 4 m/s Vi' = ? V2 = 7 m/s mi m2 Car ma m2 In the above collision what is vi'? 0 -8 m/s O 3 m/s O-1 m/s O 5 m/s O 1 m/s O-3 m/s O 8 m/s 0-2 m/ O -5 m/s O 4 m/s 0-4 m/s 0 2 m/s
Solve for v1 and v2 given conservation of kinetic
energy
Search this book n, in this case, reads Pi = Pf | M2V2,i = mjº18+ m2V2,f. ergy reads K; = K mavzi = {m101s+ mauz.s. lons in two unknowns. The algebra is tedious but not terribly difficult; you de (mi-m2)+1,i+2m2 U2, mi + m2 (m2-mi U2,+2m, V1, mi +m2 U2f numbers, we obtain V1,8 V2,1 = = 2.22 m -0.28 m
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