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