Post on 05-Jan-2016
description
23.0 m/s
Conservation of momentumConservation of momentum ProjectileProjectile
mv + mv = mv
5.5g22.6 g
1.5m
2.5m
Velocity after the collision Velocity after the collision is the horizontal velocity is the horizontal velocity
for the projectilefor the projectilev
VERT
vi = 0
d = 1.5
a = 9.8
t = ?
d = vit + 1/2at 2
1.5 = 1/2(9.8)t2
t = .553 s
HOR
v =d
t
v =2.5
.553
v =4.52 m/s
5.5v = (5.5+22.6)(4.52)
5.5v = 127
v = 23.0 m/s
28.7 s
Find the speed the student moves using conservation of momentum
mv = mv + mvmv = mv + mv
0 = 2.6(-5) + 74.5v
v = .174 m/s
Student will keep sliding at constant speed because there is no friction
t = 28.7 sv = d
t .174 =
5
t
14.5 m/s north
BEFORE AFTER
V? 10m/s2550 1550
5.22 m/s
mv + mv = mvmv + mv = mv
2550v + 1550(-10) = 4100(5.22)
2550v - 15500 = 21402
2550v = 36902
v = 14.5 m/s
14 cm
.4 .63.5
.4 .6.7 v?
mv + mv = mv + mvmv + mv = mv + mv
.4(3.5) = .4(-.7) + .6v
1.4 = - .28 + .6v
v = 2.8 cm/sThere is no friction, so bead continues at constant speed
d = 14 cmv = d
t 2.8 =
d
5
339 m/s
8 g2.5 kg
6 cm
v?
Ebottom = Etop
1/2 mv2 = mgh
.5v2 = 9.8(.06)
v = 1.08 m/sThis is the velocity of the This is the velocity of the combination immediately combination immediately
after the collisionafter the collision
BEFORE AFTER
mv + mv = mv
.008v = 2.508(1.08)
v = 339 m/s
A. .833 m/s
B. 1.17 m/s left
A. F t = mv (2.5)(.5) = 1.5v .833 m/s = v
This is the answer since it started at rest
B. B. We know the CHANGE in velocity is .883 We know the CHANGE in velocity is .883
… …it could be .883 MORE or .883 LESSit could be .883 MORE or .883 LESSSince it was moving to the left and the force was to the right, it was slowed down .833 m/s
2.0 - .833 = 1.17 m/s