2 Hints8.7—Calculate the change in momentum divide by the period in which it happened.8.22—Solve the KE for the velocity and use conservation of momentum8.36—Use cons. Of momentum to calculate the mass of the bullet-block system; plug result into conservation of energy8.35—The block-bullet system is stopped by work done by friction (DK-Wfriction=0) over the distance, s. Now use cons. Of momentum to find the velocity8.42—Part a) use Eq b) KE is proportional to v2 c) Assume a fraction in part a is x then xn is equal to 1/59,0008.46—Part a) Use equation 8.28 Part c) Use Eq. 8-308.68—(kg/bullet)*(m/s)*(bullet/s)=newtons8.70—Use cons. Of energy to solve for the initial velocity b) use conservation of energy
3 8.39A 0.15 kg glider is moving to the right on a frictionless surface with a speed of 0.8 m/s. It has a head-on collision with a 0.3 kg glider that is moving to the left with a speed of 2.2 m/s. Find the final velocity (magnitude and direction) of each glider if the collision is elastic.
4 Set It up m1= mass of glider 1=0.15 kg m2= mass of glider 2=0.3 kg vi1=initial velocity of glider 1=0.8 m/svi2=initial velocity of glider 2=-2.2 m/sCollision is elastic; therefore KE is conserved.
6 2 Solutions Using the “-” solution, v1f=-3.2 m/s v2f=-0.2 m/s Physical intuition tells us that the lighter glider (1) will not push the heavier, faster moving glider out of the way so the “-” solution is correct.
7 8-63Three identical pucks on a horizontal air table have repelling magnets. They are held together and then released simultaneously. Each as the same speed at any instant. One puck moves due west. What is the magnitude and direction of the other two pucks?
11 8.71A 0.1 kg stone rests on a frictionless horizontal surface. A bullet of mass 6.0 g traveling horizontally at 350 m/s strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 m/s.Compute the magnitude and direction of the velocity of the stone after it is struckIs the collision perfectly elastic?
13 Conservation of Mom. In both x and y directions
14 Part B) Compare KEbefore to KEafter KEbefore = ½ *(0.006)*3502= 367 JKEafter = ½ *(0.006)* ½ *(.1)*25.82KEafter =221 JKEbefore not equal KEafter so not perfectly elastic
15 8.96A 20 kg projectile is fired at an angle of 600 above the horizontal with a speed of 80 m/s. At the highest point in the trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. Ignore air resistance.How far from the point of firing does the other fragment strike if the terrain is level?How much energy is released during the explosion?
16 From ProjectilesWe know we must break initial velocity into horizontal and vertical componentsvh=v*cos(600)=0.5*80m/s=40 m/svv=v*sin(600)=0.86*80 m/s =69 m/sIn the vertical direction at highest point:v(t)=0=-g*t+vv or t=vv/g=69/9.8=7 m/sy(t)=- ½ *g*t2+vv*t= -0.5*9.8*49+69*7 =242 mx(t)=vh*t=40*7=280 m
17 Draw Before and After Before After px1 py1=0 px px2=0 py=0 py2=0 If “2” is just dropping with zero speed, thenY direction: 0=py1+py2 but py2=0 so py1=0
18 Draw Before and After Before After px1 py1=0 px px2=0 py=0 py2=0 If “2” is just dropping with zero speed, thenX direction: px=px1+px2 but px2=0 so px1=px0.5*m*vx2=mvh which means thatVx2=2*vh=80 m/s
19 Finishing part a) Now, it takes 2*7 seconds for objects to hit ground. Mass “2” hits ground at 280 m from starting point.Mass “1” has now accelerated at this point to 2*vhX(t)=80*7+280=840 m from launch point
20 Part b) KEafter-Kbefore=Energy of explosion KEafter= ½ *(m/2)*(2*vh)2= mvh2KEbefore= ½ mvh2Energy then is ½ *m*vh2=0.5*20*40^2Energy=16000 J