A bullet of mass 4.00 grams speeds horizontally along thepositive x-axis with a

Question

A bullet of mass 4.00 grams speeds horizontally along thepositive x-axis with a velocity whose x component is 21.0 m/s. Itcollides with a block of mass 0.230  kilograms which isat rest on a smooth horizontal surface. The bullet completelypasses through the block and emerges on the other side. After thebullet emerges, the block has an x component of velocity 0.291 m/s.Ignore any change of mass of the block resulting from the bullet'spassage through it. What is the x component of the velocity of the center of massof the system after the collision? A bullet of mass 4.00 grams speeds horizontally along thepositive x-axis with a velocity whose x component is 21.0 m/s. Itcollides with a block of mass 0.230  kilograms which isat rest on a smooth horizontal surface. The bullet completelypasses through the block and emerges on the other side. After thebullet emerges, the block has an x component of velocity 0.291 m/s.Ignore any change of mass of the block resulting from the bullet'spassage through it. What is the x component of the velocity of the center of massof the system after the collision?

Explanation / Answer

Data
mass of bullet m b = 4.00 g = 0.004kg mass of block M b = 0.23kg velocity of bullet v b1x = 21.0m/s velocity of block Vb2x = 0.291m/s   According to conservation of momentum mbvb1x +MbVb1x = mbvb2x +MbVb2x mbvb1x =mbvb2x + MbVb2x ( mbvb1x -MbVb2x ) / mb =vb2x    vb2x = ( mbvb1x- MbVb2x ) / mb           = (0.004kg x 21.0m/s - 0.23kg x 0.291 m/s ) / 0.004kg ( mbvb1x -MbVb2x ) / mb =vb2x    vb2x = ( mbvb1x- MbVb2x ) / mb           = (0.004kg x 21.0m/s - 0.23kg x 0.291 m/s ) / 0.004kg           =(0.084kg.m/s -0.06693 kg.m/s )/0.004kg
          =4.2675m/s

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