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A 7.2 kg block with a speed of 10 m/s collides with a 19 kg block...

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Physics

A 7.2 kg block with a speed of 10 m/s collides with a 19 kgblock that has a speed of 5.4 m/s in the same direction. After thecollision, the 19 kg block is observed to be traveling in theoriginal direction with a speed of 5.4 m/s. (a)What is the velocity of the 7.2 kg block immediately after thecollision?(b) By how much does the total kineticenergy of the system of two blocks change because of the collision?(c) Suppose, instead, that the 19 kg block ends upwith a speed of 4.7 m/s. What then is the change in the totalkinetic energy?

Answer & Explanation Solved by verified expert
4.2 Ratings (689 Votes)

Where m1 = Mass of first body ,=7.2kg.
m2 = Mass of second body,=19kg.
u1 = Initial velocity of first body,=10m/s
u2 = Initial velocity of second body,=5.4m/s
v1 = Final velocity of first body,=?
v2 = Final velocity of second body=5.4m/s.

it is a elastic collision.

so,m1u1+m2u2=m1v1+m2v2.

(7.2)(10)+(19)(5.4)=(7.2)(v1)+(19)(5.4).

174.6=7.2v1+102.6.

   7.2v1=174.6-102.6

     :.v1=72/7.2

    :.v1=10m/s.

2.change in kinetic energy is given by=m1v12/2+m2v22/2.

                                                         =[(7.2)(102)/2]+[(19)(5.42)/2].

                                     :.K.E.=360+277.02

                                    :.K.E=637.02J.

3.change in kinetic energy is=m1v12/2+m2v22/2.

                                           =[(7.2)(102)/2]+[(19)(4.72)/2].

                                :.K.E=360+209.855

                             :.K.E.=569.855J.
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