m- m1 V2 Vi 01 1 02 V2 3
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Q: In the figure, mass m1=10.0 kg is moving at a speed of Vi1 undergoes a collision with mass m2 = 7.00…
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Two objects are moving towards each other with velocities V→1 and V→2 on a frictionless surface. They collide completely inelastically and move together after the collision with a velocity |V→3|=16.5m/s as shown below.If the ratio of the masses is m1/m2=1.7 and the angles are θ1=34o and θ2=45o, what is the magnitude of V→2 in units of m/sm/s? Express your answer using one decimal place.
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- A 2.10 kg "particle" traveling with velocity = (3.7 m/s )i collides with a 4.70 kg "particle" traveling with velocity V = (1.7 m/s )j. The collision connects the two particles. What then is their velocity in (a) unit-vector notation and as (b) a magnitude and (c) a positive (counterclockwise) angle measured from the +x axis? (a) Number i+ į Units (b) Number Units (c) Number UnitsA proton with an initial speed of 1.75 ✕ 108 m/s in the +x direction collides elastically with another proton initially at rest. The first proton's velocity after the collision is 1.416 ✕ 108 m/s at an angle of 36.0° with the +x-axis. What is the velocity (magnitude and direction) of the second proton after the collision? Magnitude(m/s): Direction( degree counterclockwise from the +x-axis):A 10.50-g bullet is fired into a stationary block of wood having mass m = 4.980 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.6090 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.) m/s
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- Two balls of clay, with masses M1 = 0.39 kg and M2 = 0.42 kg, are thrown at each other and stick when they collide. Mass 1 has a velocity v1 = 3.4i m/s and mass 2 has a velocity of v2 = 1.75j m/s.a. Write an expression for the initial momentum of the system, P, in terms of the masses M1 and M2, the magnitudes of the velocities v1 and v2 and the unit vectors i and j.A 9.7-g bullet is fired into a stationary block of wood having mass m = 5.02 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.603 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.)A mass of m1= 4 kg moving with a velocity of v1=(2 m/s)ˆi (positive x−direction) collides with another mass, m2 moving at v2= (−4 m/s)ˆi (negative x−direction). The masses collide and stick together, moving off with a velocity of v= (−2 m/s)ˆi. What is m2? A)m2= 4 kg B)m2= 8 kg C)m2= 12 kg D)m2= 16 kg
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