Two people of unequal mass are initially standing still on ice with negligible f

Question

Two people of unequal mass are initially standing still on ice with negligible friction. They then simultaneously push each other horizontally. Afterward, which of the following is true? The kinetic energies of the two people are equal. The speeds of the two people are equal. The momenta of the two people are of equal magnitude. The center of mass of the two-person system moves in the direction of the less massive person. The less massive person has a smaller initial acceleration than the more massive person.

Explanation / Answer

Mass of the two people are m and M .

From problem , m is not equal to M

Initial speeds of two people is zero.So, total momentum before push = 0

From law of conservation of linear momentum ,total momentum after push = 0

m v + M(-V) = 0 Here v and -V are the velocities of mass m and M after push.

after push m and M moves in opposite in direction.one velocity is positive and other velocity is negative.

mv = MV

Momentum of m after push = momentum of M after push.

i.e., option(c) is correct

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