C with speed 2 v d with speed 2 v e none of the above

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(c) with speed 2 v . (d) with speed 2 v . (e) none of the above 7
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Again we must use the concept of conservation of momentum. The prob- lem states that the first two fragments split into orthogonal directions, so we are free to define the positive ˆ x direction as the direction of the fragment of mass 3 m and the positive ˆ y direction as the direction of the fragment of mass 4 m . Another key concept is the conservation of mass. The system had a total mass of 8 m prior to the explosion, thus it must have a total mass of 8 m after the collision. Therefore, the mass of the third fragment must be m so that the three fragments have a total mass of 8 m . The total momentum of the system before the explosion, ~ p 0 , is 0. ~ p 0 8 m~v 8 m 0 0 . This is a vector equation, so a 0 means 0 in each direction. So we must have that the momentum of the fragments after the collision also sum to 0 in each direction. It is often useful to break up problems about momenta into components. Summing the momenta in the ˆ x direction: p f,system,x 0 p f, 3 m,x p f, 4 m,x p f,m,x 3 mv 0 mv f,m,x . This gives us the value of the third fragments velocity in the ˆ x direction in terms of v , v f,m,x 3 v . Similarly for the ˆ y component of the momentum: p f,system,y 0 p f, 3 m,y p f, 4 m,y p f,m,y 0 4 mv mv f,m,y . This gives us the value of the third fragments velocity in the ˆ y direction in terms of v , v f,m,y 4 v . The speed of a particle is defined as the length of the velocity vector, so using the equation for the magnitude of a vector: ~v f,m v 2 f,m,y v 2 f,m,x 4 v 2 3 v 2 5 v. This is choice a. Remember, this was a special case where the total momenta in each direction was 0. However, if for instance, the initial mo- menta of the system were 3 mv ˆ x , i.e. ~ p 0 3 mv ˆ x , then we would have set p f,system,y 0, and p f,system,x 3 mv . 8
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6. A 2.0 kg toy rocket is rigidly attached to one end of a horizontal 5.0 m- length rod of negligible mass on a frictionless table. The rod is free to rotate about the opposite end, which is fixed to the table, so the rocket can move in a circle centred at the opposite end of the rod. As shown in the figure, the rocket is aligned so that its 30 N-thrust points horizontally, at 30 to the rod. If the rocket starts moving from rest, what is its angular velocity after it completes exactly 2 full turns? 5.0 m rocket 30 (a) 6.1 rad/s 52 (b) 8.1 rad/s (c) 58 rad/s (d) 46 rad/s (e) 21 rad/s Here we have a rocket attached to a rod such that rocket’s thrust will be directed 30 degrees behind the rod. The first step is again to choose a coordinate system, for circular motion, it is typically most convenient to use polar coordinates. We will define the origin as the pivot point between the rod and table, the ˆ θ direction is along the circle traced out by the rocket and the ˆ r direction pointing outwards from origin along the rod. This is completely equivalent to breaking up the force in terms of the tangential and radial components, where ˆ θ is the tangential component and ˆ r is the radial component. We can
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