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Determine the velocity of the m1 block just before the collision. (Use positive sign if the motion is to the right, negative if it is to the left.)

Determine the velocity of the m2 block just after the collision. (Use positive sign if the motion is to the right, negative if it is to the left.)
Determine the velocity of the m1 block just after the collision. (Use positive sign if the motion is to the right, negative if it is to the left.)
Determine the maximum height to which m2 rises after the collision.
Determine the maximum height to which m1 rises after the collision.

Explanation / Answer

1). The velocity of m1 just before impact is found from
Pe=Ke
m1gh=(1/2) m1 U1^2
The velocity of the m1 block just before the collision U1=sqrt(2*9.8*5.08)=9.98m/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
solving these equations we get

2). The velocity of the m2 block just after the collision

  V2= 2m1U1/(m1+m2)

=2(1.95)(9.98)/(1.95+4.4.1)

=6.118m/s

3). The velocity of the m1 block just after the collision

V1=(m1-m2)U1/(m1+m2) and since U1=sqrt(2 gh)= 7.2 m/s
V1=(1.95 - 4.41) *9.98/(1.95 +4.41)
the velocity of the m1 block just after the collisionV1= -3.86m/s



Again Ke(1/2)mV^2=mgh
h2=(4.41)(6.12)^2 /9.8

= 165.065/9.8

= 16.84m

Time of fall =time of horizontal travel. Horizontal velocity is constant
t=sqrt(2 h2 / g)
t= sqrt( 2 x 16.84 / 9.81)=3.34 s
S= V2 t= 6.118*3.34=
4). the maximum height to which m2 rises after the collision S=20.43m

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

5). the maximum height to which m1 rises after the collision S= V1t=3.86 x 3.34= 12.89 m

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