Question Part Points Submissions Used Question Part Points Submissions Used Ques

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Question Part Points Submissions Used

Question Part Points Submissions Used

Question Part Points Submissions Used

Question Part Points Submissions Used Question Part Points Submissions Used m1 = 0.450 kg h1 = 3.10 m m2 = 0.900 kg (a) Determine the velocities of the two objects just after the collision. (Enter the magnitude of the velocity.) v1 =  m/s v2 =  m/s
(b) How high up the track does the 0.450-kg object travel back after the collision?
m

(c) How far away from the bottom of the table does the 0.900-kg object land, given that the height of the table is h2 = 1.95 m?
m

(d) How far away from the bottom of the table does the 0.450-kg object eventually land?
m h2 = 1.95 m? Question Part Points Submissions Used m

Explanation / Answer

a) The velocity of m1 just before impact is found from
Pe=Ke
m1gh=(1/2) m1 U1^2
U1=sqrt(2 gh)=7.8m/s

Conservation of momentum
m1U1 +m2U2=m1V1 + m2V2
It helps to know that U2= 0

m1U1 = m1V1 + m2V2
also
What about conservation of kinetic energy? Ke=(1/2) mV^2
Ke1+Ke2=Ke1' + Ke2' again Ke2=0 and 1/2 term can be dropped

m1U1^2= m1V1^2 + m2V2^2 (See ref 1)
solving these equations we get

V1=(m1-m2)U1/(m1+m2) and since U1=sqrt(2 gh)= 7.2 m/s
V1=(0.450 - 0.900) (7.8)/(0.450 +0.900)
V1=-2.6m/s

V2= 2m1U1/(m1+m2) = 5.2 m/s

b) Again Ke(1/2)mV^2=mgh
h=(1/2)V^2 /g= 0.5 x (2.6)^2 /9.81=.342 m

c) Time of fall =time of horizontal travel. Horizontal velocity is constant
t=sqrt(2 h2 / g)
t= sqrt( 2 x 1.95 / 9.81)=0.6305 s
S= V2 t= 2m1U1/(m1+m2) t=
S=[2 x 0.450 x 7.8/(0.45 +0.900) ] 0.6305 =
S=3.278 m

d) The horizontal velocity will be the same in magnitude as it was after m1 colided with m2. The time of flight is the sam efor both of them so

S= V1t= 2.6 x 0.6305= 1.63 m

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