Presentation on theme: "Hints and Examples from Chapter 8. Hints 8.7—Calculate the change in momentum divide by the period in which it happened. 8.22—Solve the KE for the velocity."— Presentation transcript:
Hints and Examples from Chapter 8
Hints 8.7—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 momentum 8.36—Use cons. Of momentum to calculate the mass of the bullet-block system; plug result into conservation of energy 8.35—The block-bullet system is stopped by work done by friction ( K-W friction =0) over the distance, s. Now use cons. Of momentum to find the velocity 8.42—Part a) use Eq b) KE is proportional to v 2 c) Assume a fraction in part a is x then x n is equal to 1/59, —Part a) Use equation 8.28 Part c) Use Eq —(kg/bullet)*(m/s)*(bullet/s)=newtons 8.70—Use cons. Of energy to solve for the initial velocity b) use conservation of energy
8.39 A 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.
Set It up m 1 = mass of glider 1=0.15 kg m 2 = mass of glider 2=0.3 kg v i1 =initial velocity of glider 1=0.8 m/s v i2 =initial velocity of glider 2=-2.2 m/s Collision is elastic; therefore KE is conserved.
Cons of E and Mom.
2 Solutions Using the “-” solution, v 1f =-3.2 m/s v 2f =-0.2 m/s Using the “+” solution, v 1f =0.8 m/s v 2f =-2.2 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.
8-63 Three 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?
Draw It! v v v
Conservation of momentum in the north-south direction says: v v v v*sin v*sin v*sin = v * sin
8.71 A 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. a) Compute the magnitude and direction of the velocity of the stone after it is struck b) Is the collision perfectly elastic?
Draw “Before” and “After” m 1 =6 g v 1i =350 m/s m 2 =100 g v 2i =0 m 1 =6 g v 1f =250 m/s m 2 =100 g v 2f =? v 2f *sin
Conservation of Mom. In both x and y directions
Part B) Compare KE before to KE after KE before = ½ *(0.006)*350 2 = 367 J KE after = ½ *(0.006)* ½ *(.1)* KE after =221 J KE before not equal KE after so not perfectly elastic
8.96 A 20 kg projectile is fired at an angle of 60 0 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. a) How far from the point of firing does the other fragment strike if the terrain is level? b) How much energy is released during the explosion?
From Projectiles We know we must break initial velocity into horizontal and vertical components v h =v*cos(60 0 )=0.5*80m/s=40 m/s v v =v*sin(60 0 )=0.86*80 m/s =69 m/s In the vertical direction at highest point: v(t)=0=-g*t+v v or t=v v /g=69/9.8=7 m/s y(t)=- ½ *g*t 2 +v v *t= -0.5*9.8*49+69*7 =242 m x(t)=v h *t=40*7=280 m
Draw Before and After pxpx p y =0 p x1 p y2 =0 p x2 =0 If “2” is just dropping with zero speed, then Y direction: 0=p y1 +p y2 but p y2 =0 so p y1 =0 p y1 =0
Draw Before and After pxpx p y =0 p x1 p y2 =0 p x2 =0 If “2” is just dropping with zero speed, then X direction: p x =p x1 +p x2 but p x2 =0 so p x1 =p x 0.5*m*v x2 =mv h which means that V x2 =2*v h =80 m/s p y1 =0
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*v h X(t)=80*7+280=840 m from launch point
Part b) KE after -K before =Energy of explosion KE after = ½ *(m/2)*(2*v h ) 2 = mv h 2 KE before = ½ mv h 2 Energy then is ½ *m*v h 2 =0.5*20*40^2 Energy=16000 J