4. A bullet with mass = 0.5 kg is shot to the side with a speed of 200 m/s. It h

Question

4. A bullet with mass = 0.5 kg is shot to the side with a speed of 200 m/s. It hits a wooden block with mass = 1.5 kg and passes through the block. As it exits the block, the bullet has a speed of 80 m/s. The block is initially at rest and is on a frictionless surface. a. Find the speed of the block after the bullet leaves it. b. Which object, the bullet or the block, got the greatest impulse as a result of the collision, and what is that impulse? At a busy intersection, a 1500 kg blue car is traveling south. At the same time, a 2000 kg red pickup is traveling west. Unfortunately, they collide, They stick together and end up with a total momentum of 8000 kgm/s directed 600 west of south. c. Find the speed of each car before the collision.

Explanation / Answer

4)

initial momentum Pi = m1*u1 = 0.5*200

final momentum = m1*v1 + m2*v2 = (0.5*80)+(1.5*v2)

from momentum conservation

Pf = Pi

(0.5*80)+(1.5*v2) = 0.5*200

v2 = 40 m/s

+

(b)

for bullet impulse = J1 = m1*(V1-u1) = 0.5*(80-200) = -60 kg m/s

for block J2 = m2*(v2-u2) = 1.5*(40-0) = 60 kgm/s

both has same impulse

+++++++

c)

u1x = 0           u1y = -u1 j

u2x = -u2          u2y = 0

after collision

Pfx = -8000*cos60

Pfy = -8000*sin60

from momentum conservation

Pix = Pfx

(1500*u1x) + (2000*u2x) = -8000*cos60

0 - (2000*u2x) = -8000*cos60

u2x = 2 m/s

=====

Piy = Pfy

-(1500*u1y) + 0 = -8000*sin60

u1y = 4.62 m/s

speed of 1500 kg car = 4.62 m/s

speed of 2000 kg car = 2 m/s

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