1) A 5.00-g bullet is fired horizontally into a 1.50-kg wooden block resting on

Question

1) A 5.00-g bullet is fired horizontally into a 1.50-kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.30 m along the surface before stopping. What was the initial speed of the bullet?

2) On a very muddy football field, a 120-kg linebacker tackles an 88-kg halfback. Immediately before the collision, the linebacker is slipping with a velocity of 8.7 m/s north and the halfback is sliding with a velocity of 6.9 m/s east. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision? Magnitude___________m/s Direction___________onorth of east

3) Two vehicles are approaching an intersection. One is a 2850-kg pickup traveling at 15.1 m/s from east to west (the ?x-direction), and the other is a 1475-kg sedan going from south to north (the +y-direction at 20.9 m/s).

(a) Find the x- and y-components of the net momentum of this system.

Explanation / Answer

Since you are only allowed to ask one question at a time, and we are only allowed to answer one at a time when only offering 300 points, I will answer question 1 for you. Repost the others separately and I can answer them as well.

Here is number 1)

The initial KE of the bullet-block combo is lost by the work due to friction

KE = W

.5mv2 = umgd (mass cancels)

.5(v)2 = (.2)(9.8)(.3)

v = 1.084 m/s

Then use that and conservation of momentum to find the initial speed

mv + Mv = (M + m)v

(.005)(v) + 0 = (1.505)(1.084)

v = 326.4 m/s

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.