.A 4.70 g bullet is fired horizontally at two blocks sitting at rest on a fricti
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Question
.A 4.70 g bullet is fired horizontally at two blocks sitting at rest on a frictionless table. The bullet passes through block 1 (mass 1.20 kg) and becomes embedded in block 2 (mass 1.80 kg). The blocks end up with speeds v1 = 0.630 m/s and v2 = 1.40 m/s. Neglect the mass removed from the first block by the bullet.
Find the speed of the bullet as it leaves block 1.
Find the speed of the bullet as it enters block 1
Two 1.80 kg bodies, A and B, collide. The velocities before the collision are Va=12i+30j m/s and Vb=-10i+15j. Immediately after the collision block A has Vaf=-5i+18j.
Wat is the velocity of block B immediately after the collision?
What is the change in the total kinetic energy because of the collision?
4.
A helicopter lifts a 72 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is 0.10g.
How much work is done on the astronaut by the force from the cable?
How much work is done on the astronaut by the gravitational force?
What is the astronaut’s kinetic energy just before she reaches the helicopter?
What is the speed of the astronaut just before she reaches the helicopter?
A 18.0 kg block is being pushed up a ramp by a force FP = 150 N acting horizontally. The ramp is 5.0 m long, inclined at 32° and the coefficient of kinetic friction between the block and the ramp is k = 0.10.
Determine the work done on the block by FP.
Determine the work done on the block by the gravitational force.
Determine the work done on the block by the normal force.
Determine the work done on the block by the frictional force.
Determine the speed of the block at the top of the ramp if it started at the bottom of the ramp with a speed of 0.25 m/s.
A moving object of mass 0.20 kg experiences two forces, F1=(1.5i-0.8j+0.7k)N and F2=(-0.7i+1.2j)N The resulting displacement of the object due to the forces is given by d=(8.0i+6.0j+5.0k)m. Determine the amount of work done on the object by the two forces.
Explanation / Answer
Problem -1
Work backwards to find the speed of the bullet before (or as) it strikes and embeds itself into the second block. The law of conservation of momentum says that the initial momentum is equal to the final momentum. In equation form :
mv(i) + mv(i) = (m + m)v(f)
The second block is initially at rest, so its momentum is zero, and the equation becomes :
mv(i) = (m + m)v(f)
Solved for v(i) :
v(i) = (m + m)v(f) / m
= [(0.0047kg + 1.80kg)(1.40m/s)] / 0.0047kg
= 537.6m/s
The speed of the block as it leaves block 1 is 537.6m/s
For its speed as it enters block 1 :
mv(i) + mv(i) = mv(f) + mv(f)
In this case, m is block 1, and its initial momentum is zero. The equation above, then, becomes :
mv(i) = mv(f) + mv(f)
v(i) = [mv(f) + mv(f)] / m
= [(0.0047kg)(537.6m/s) + (1.20kg)(0.630m/s)] / 0.0047kg
= 699.15m/s
The speed of the bullet before it strikes block 1 is 699.15 m/s
Problem-2
Applying conservationof momentum
P inital= P final
mVa + mVb= mVaf+ mVbf
1.80 [ 12i+30j ] +1.80 [-10i+15j] = 1.80 [ -5i+18j] +1.80 * Vbf
[ 12i+30j ] + [-10i+15j] = [ -5i+18j] +Vbf
2i+45j= [ -5i+18j] +Vbf
7i - 27 j= Vbf
The velocity of block B immediately after the collision is [7i - 27 j]
Ke initial= 1/2mVa2+ 1/2mVb2 = [1/2*1.80* 32.312]+ [1/2* 1.80*18.032]= 939.5+292.6= 1232.1 J
Ke initial= 1/2mVaf2+ 1/2mVbf2 = [1/2*1.80* 18.682]+ [1/2* 1.80*27.892]= 314.05+389.9= 703.95J
Change in the total kinetic energy because of the collision = [1232.1J- 703.95J] = 528.15J
Problem-3
work is done on the astronaut by the force from the cable= mg*d= 72*9.8*15=10584 J [upwards]
work is done on the astronaut by the gravitational force=10584 J [downwards]
Astronaut’s kinetic energy just before she reaches the helicopter= Pe= mgh= 10584 J
speed of the astronaut just before she reaches the helicopter
1/2mv2= 10584 J
v2= 10584*2/ 72= 294
v= 17.15 m/s
Speed of the astronaut just before she reaches the helicopter is 17.15 m/s
Problem- 4
Work done on the block by FP= 150N *5m* sin[32o] = 397.44 J
Work done on the block by the gravitational force, Fg=150N *5m* cos[32o] = 636.04 J
Work done on the block by the normal force, Fn=150N *5m* cos[32o] = 636.04 J
Work done on the block by the frictional force, Ff=k*Fn = 0.10*636.04 J = 63.6J
speed of the block at the top of the ramp if it started at the bottom of the ramp with a speed of 0.25 m/s
Vi=0.25 m/s
a= [gsintheta-k*g*cos theta] =9.8*sin[320]-0.10*9.8*cos[320] = 5.19- 0.83= 4.36 m/s2
d= 5m
using the kinematic equation; Vf2= Vi2+2ad
Vf2= 0.252+ [2*4.36*5] = 43.65
Vf= 6.61 m/s
Speed of the block at the top of the ramp is 6.61 m/s
Problem-5
F= F1+ F2= (1.5i-0.8j+0.7k)N + (-0.7i+1.2j)N = (0.8i+0.4j+0.7k)N
d=(8.0i+6.0j+5.0k)m
W= F.d= (0.8i+0.4j+0.7k)N .(8.0i+6.0j+5.0k)m = 6.4+2.4+3.5 = 12.3 J
The amount of work done on the object by the two forces is 12.3 J
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