A 3.1 kg block with a speed of 5.4 m/s collides with a 6.2 kg block that has a s

Question

A 3.1 kg block with a speed of 5.4 m/s collides with a 6.2 kg block that has a speed of 3.6 m/s in the same direction. After the collision, the 6.2 kg block is observed to be traveling in the original direction with a speed of 4.5 m/s. (a) What is the velocity of the 3.1 kg block immediately after the collision? (b) By how much does the total kinetic energy of the system of two blocks change because of the collision? (c) Suppose, instead, that the 6.2 kg block ends up with a speed of 7.2 m/s. What then is the change in the total kinetic energy?

Explanation / Answer

befroe collision,

mass of the first block, m1=3.1 kg,

speed,u1=5.4 m/sec

mass of the second block, m2=6.2 kg,

speed, u2=3.6 m/sec

after collsion,

speed of the m1 is v1

speed of the m2 is v2 = 4.5 m/sec

a)

by using law of conservation of momentum,

m1*u1+m2*u2=m1*v1+m2*v2

3.1*5.4+6.2*3.6 = 3.1*v1+6.2*4.5

===> v1=3.6 m/sec

b)

change in dK=K1-K2

=(1/2*m1*u1^2+1/2*m2*u2^2) - (1/2*m1*v1^2+1/2*m2*v2^2)

=(1/2*3.1*5.4^2+1/2*6.2*3.6^2) - (1/2*3.1*3.6^2+1/2*6.2*4.5^2)

=2.511 J

c)

if v2=7.2 m/sec

then,

change in dK=K2-K1

=(1/2*m1*v1^2+1/2*m2*v2^2) - (1/2*m1*u1^2+1/2*m2*u2^2)

=(1/2*3.1*3.6^2+1/2*6.2*7.2^2) - (1/2*3.1*5.4^2+1/2*6.2*3.6^2)

=95.418 J

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