Consider a frictionless track as shown in the figure below. A block of mass m_1

Question

Consider a frictionless track as shown in the figure below. A block of mass m_1 = 5.00 kg is released from A. It makes a head on elastic collision at B with a block of mass m_2 = 10.0 kg that is initially at rest. Calculate the maximum height to which m_1 rises after the collision. A car of mass m traveling at speed v crashes into the rear of a truck of mass 2m that is at rest and in neutral at an intersection. If the collision is perfectly inelastic, what is the speed of the combined car and truck after the collision? 2v v v/2

Explanation / Answer

Q7.

speed of m1 just before collision=sqrt(2*g*height)

=9.8995 m/s

let speed of m1 after collision is v1 to the left and speed of m2 after collision is v2 to the right

conserving momentum:

momentum before collision=momentum after collision

==>5*9.8995=-5*v1+10*v2

==>-v1+2*v2=9.8995

==>v1=2*v2-9.8995...(1)

conserving total kinetic energy:

0.5*5*9.8995^2=0.5*5*v1^2+0.5*10*v2^2

==>v1^2+2*v2^2=9.8995^2

using equation 1,

4*v2^2+9.8995^2-2*2*9.8995*v2+2*v2^2=9.8995^2

==>6*v2^2-2*2*9.8995*v2=0

==>v2=6.6 m/s

then v1=2*v2-9.8995=3.3 m/s

then maximum height achieved by m1=v1^2/(2*g)

=0.5556 m

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