A 3.2-kg block is traveling in the x direction at 4.6 m/s, and a 1.1-kg block is

Question

A 3.2-kg block is traveling in the x direction at 4.6 m/s, and a 1.1-kg block is traveling in the +x direction at 3.3 m/s.

(a) Find the velocity vcm of the center of mass. m/s

(b) Subtract vcm from the velocity of each block to find the velocity of each block in the center-of-mass reference frame. 3.2-kg block m/s 1.1-kg block m/s

(c) After they make a head-on elastic collision, the velocity of each block is reversed (in the center-of-mass frame). Find the velocity of each block in the center-of-mass frame after the collision. 3.2-kg block m/s 1.1-kg block m/s

(d) Transform back into the original frame by adding vcm to the velocity of each block. 3.2-kg block m/s 1.1-kg block

m/s

(e) Check your result by finding the initial and final kinetic energies of the blocks in the original frame and comparing them.

Explanation / Answer

here,

m1 = 3.2 kg has v1 = - 4.6 m/s

m2 = 1.1 kg has v2 = 3.3 m/s

(a)

speed of centre of mass, v= m1v1 + m2v2 /( m1 + m2)

v = ( 3.2 *(-4.6) + 1.1 * 3.3 ) /( 3.2 +1.1)

v = 2.47 m/s in -x direction

(b)

the velocity of 3.2 kg block in the center-of-mass reference frame , vc1 = v1 - v = 2.13 m/s in - x direction

the velocity of 1.1 kg block in the center-of-mass reference frame , vc2 = v2 - v

vc2 = 3.57 m/s in x direction

(c)

using conservation of momentum

m1u1 + m2u2 = m1v1 + m2v2

3.2 *(-2.13) + 1.1 * 3.57 = 3.2 * v1 + 1.1 * v2.......(1)

and

using conservation of kinetic energy

0.5 * m1*u1^2 + 0.5 m2*u2^2 = 0.5 * m1*v1^2 + 0.5 * m2* v2^2

3.2 *(2.13)^2 + 1.1 * 3.57^2 = 3.2 * v1^2 + 1.1 * v2^2 ...(2)

from (1 ) and (2)

v1 = 0.79 m/s

v2 = - 4.91 m/s

(d)

vc1 = v1 + vm

vc1 = - 1.68 m/s

vc2 = v2 + vm

vc2 = - 7.48 m/s

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