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A 15.2 g bullet is shot into a 4763 g wooden block standing on a frictionless surface,...

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w410502 | Student, College Freshman | eNotes Newbie

Posted October 8, 2011 at 3:06 PM via web

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A 15.2 g bullet is shot into a 4763 g wooden block standing on a frictionless surface, the acquires a speed of 1.74 m/s. What was the velocity of the bullet initially?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted October 8, 2011 at 6:17 PM (Answer #1)

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According to the conservation of linear momentum, the total initial momentum in a closed system is equal to the total final momentum if there is are no external forces. Here the surface that the block stands on is frictionless. There is no kinetic energy converted to heat.

The mass of the bullet is 15.2 g. Let its initial speed be S. The mass of the wooded block is 4763 g and it is has a speed of 0 m/s . After the bullet strikes the block, it is embedded in it and the combined system has a speed of 1.74 m/s. The mass of combined system is 4778.2 g.

Equating the initial momentum and final momentum

15.2*S = 4778.2*1.74

=> S = 4778.2*1.74/15.2

=> S = 546.97

The bullet was traveling at 546.97 m/s before it struck the wooden block.

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