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Physics Solution Manual for 1100 and 2101

# We also know that their accelerations have the same

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Unformatted text preview: e plate breaks, we have, −m1 ( 3.00 m/s ) sin 25.0° + m2 (1.79 m/s ) cos 45.0° = 0 (1) m1 ( 3.00 m/s ) cos 25.0° + m2 (1.79 m/s ) sin 45.0° − (1.30 kg ) ( 3.07 m/s ) = 0 Subtracting (2) from (1), and noting that cos 45.0º = sin 45.0º, gives (2) m1 = 1.00 kg . Substituting this value into either (1) or (2) then yields m2 = 1.00 kg . 24. REASONING We will divide the problem into two parts: (a) the motion of the freely falling block after it is dropped from the building and before it collides with the bullet, and (b) the collision of the block with the bullet. During the falling phase we will use an equation of kinematics that describes the velocity of the block as a function of time (which is unknown). During the collision with the bullet, the external force of gravity acts on the system. This force changes the momentum of the system by a negligibly small amount since the collision occurs over an extremely short time interval. Thus, to a good approximation, the sum of the external forces acting on the system during the collision is negligible, so the linear momentum of the system is conserved. The principle of conservation of linear momentum can be used to provide a relation between the momenta of the system before and after the collision. This relation will enable us to find a value for the time it takes for the bullet/block to reach the top of the building. SOLUTION Falling from rest (v0, block = 0 m/s), the block attains a final velocity vblock just before colliding with the bullet. This velocity is given by Equation 2.4 as vblock { Final velocity of block just before bullet hits it = v0, block + a t 13 2 Initial velocity of block at top of building where a is the acceleration due to gravity (a = −9.8 m/s2) and t is the time of fall. The upward direction is assumed to be positive. Therefore, the final velocity of the falling block is vblock = a t (1) 358 IMPULSE AND MOMENTUM During the collision with the bullet, the total linear momentum of the bullet/block system is conserved, so we have that m v (144+ m )3 4 2444 bullet block f = mbullet vbullet + mblock vblock 144424444 4 3 (2) Total linear momentum before collision Total linear momentum after collision Here vf is the final velocity of the bullet/block system after the collision, and vbullet and vblock are the initial velocities of the bullet and block just before the collision. We note that the bullet/block system reverses direction, rises, and comes to a momentary halt at the top of the building. This means that vf , the final velocity of the bullet/block system after the collision must have the same magnitude as vblock, the velocity of the falling block just before the bullet hits it. Since the two velocities have opposite directions, it follows that vf = −vblock. Substituting this relation and Equation (1) into Equation (2) gives (m bullet + mblock ) ( −a t ) = mbullet vbullet + mblock ( a t ) Solving for the time, we find that t= − mbullet vbullet a ( mbullet + 2mblock ) = − ( 0.015 kg ) ( +810 m/s ) ( −9.80 m/s ) ( 0.015 kg ) + 2 (1.8 kg ) 2 = 0.34 s 25. REASONING During the time that the skaters are pushing against each other, the sum of the external forces acting on the two-skater system is zero, because the weight of each skater is balanced by a corresponding normal force and friction is negligible. The skaters constitute an isolated system, so the principle of conservation of linear momentum applies. We will use this principle to find an expression for the ratio of the skater’s masses in terms of their recoil velocities. We will then obtain expressions for the recoil velocities by noting that each skater, after pushing off, comes to rest in a certain distance. The recoil velocity, acceleration, and distance are related by Equation 2.9 of the equations of kinematics. SOLUTION While the skaters are pushing against each other, the total linear momentum of the two-skater system is conserved: m1vf 1 + m2vf 2 = { 0 14 244 Total momentum 4 3 Total momentum after pushing before pushing Solving this expression for the ratio of the masses gives Chapter 7 Problems m1 m2 =− vf 2 359 (1) vf 1 For each skater the (initial) recoil velocity vf , final velocity v, acceleration a, and displacement x are related by Equation 2.9 of the equations of kinematics: v 2 = vf2 + 2a x . Solving for the recoil velocity gives vf = ± v 2 − 2ax . If we assume that skater 1 recoils in the positive direction and skater 2 recoils in the negative direction, the recoil velocities are Skater 1 vf 1 = + v12 − 2a1 x1 Skater 2 2 vf 2 = − v2 − 2a2 x2 Substituting these expressions into Equation (1) gives 2 − v2 − 2a2 x2 m1 =− = m2 v12 − 2a1 x1 2 v2 − 2a2 x2 (2) v − 2a1 x1 2 1 Since the skaters come to rest, their final velocities are zero, so v1 = v2 = 0 m/s. We also know that their accelerations have the same magnitudes. This means that a2 = −a1, where the minus sign denotes that the acceleration of skater 2 is opposite that of skater 1, since they are moving in oppo...
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