A 4.00 g bullet is moving horizontally with a velocity of + 355 m/s (see the dra

Question

A 4.00 g bullet is moving horizontally with a velocity of + 355 m/s (see the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block an inelastic collision and embeds itself in the second block. Note that both blocks are moving, after the collision with the bullet. The mass of the first block is 1150 g and its velocity is +0.550 m/s after the bullet passes through it. The mass of the second block is 1530 g. What is the velocity of the second block after the bullet embeds itself?

Explanation / Answer

Mb = 4 g ; Vb = +355 m/s ;

v1 = 0.55 m/s ; m1 = 1150g = 1.15 kg ; m2 = 1530 g = 1.53 kg

Let v2 be the required velocity of the second block.

from the conservation of linear momentum

Mb Vb = m1 v1 + (m2 + Mb)v2

v2 = [Mb Vb - m1v1/(m2 + Mb)]

v2 = [4 x 10^-3 x 355 - 1.15 x 0.55/(1.53 + 4 x 10^-3)] = 0.513 m/s

Hence, v2 = 0.513 m/s.

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