<li>A gun is fired vertically into a 1.40-kg block of wood at rest directly above it. If the bullet has a mass of 21.0 g and a speed of 210 m/s, how high will the block rise into the air after the bullet becomes embedded in it?</li>
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<p>Which formula(s) should I use to solve this problem?</p>
<p>You can split the problem into two parts. First, find the velocity of the block and bullet right after the collision via conservation of momentum. Then use a kinematics equation to find the distance it travels. You'll be best off using m1v1+m2v2=(m1+m2)*vfinal and v^2=v0^2+2a(xf-xi)</p>
<p>I calculated that the final velocity is 3.1 m/s. I plugged that into the kinematic equation, but what is my initial velocity and acceleration?</p>
<p>3.1 m/sec is the <em>initial</em> velocity of (block + bullet combination).
Use vFinal=0, vInitial= 3.1, a = -9.8 m/sec/sec in the kinematics equation.</p>
<p>Oh yea. Thanks. I forgot about gravity here.</p>
<p>Yall are only on kinematics?!</p>
<p>We're like on Power...</p>
<p>We are on momentum. Our school started on September 8th, and we didn't start our first lesson until September 13th.</p>
<ol>
<li>An atomic nucleus at rest decays radioactively into an alpha particle and a smaller nucleus. What will be the speed of this recoiling nucleus if the speed of the alpha particle is 3.8 x 10^5 m/s? Assume the recoiling nucleus has a mass 57 times greater than that of the alpha particle.</li>
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<p>This is what I started doing:</p>
<p>mava + mrvr = mava' + mrvr'
0 + 0 = ma(3.8 x 10^5 m/s) + 57mavr'</p>
<p>But then I see two variables in which I need to calculate the velocity of recoiling nucleus (vr'). Is this not the correct equation to use?</p>
<p>We did a lesson the first day...</p>
<p>Nvm. I figured it out. I just have to move one term to the other side, and cancel ma.</p>
<p>One last question that I got stuck on after attempting it.</p>
<ol>
<li>A 13-g bullet traveling 230 m/s penetrates a 2.0-kg block of wood and emerges going 170 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?</li>
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<p>This is what I did:</p>
<p>m1v1 + m2v2 = (m1 + m2)vi
(.013 kg)(230 m/s) + (2.0 kg)(0m/s) = .013 kg + 2.0 kg)vi
2.99 kg m/s + 0 m/s = 2.013 kg)vi
vi = 1.48 m/s</p>
<p>Thats as far as I got. I don't know what the vi is for, and I don't know where to go from there. Please help me.</p>
<p>Can someone help me with this please?</p>
<p>The block and the bullet have different velocities after the bullet emerges. </p>
<p>m1v1 + m2v2 = m1(v1new) + m2(v2new)
(.013 kg)(230 m/s) + (2.0 kg)(0m/s) = (.013 kg)(170 m/s) + (2.0)(v2new)
v2new = (0.013)(230-170)/2 = 0.39 m/sec</p>
<p>actually optimizerdad, the bullet is imbedded into it so you have m1v1 + m2v2 = (m1 + m2)v_f so solving for vf you get 27.4 m/s so that becomes your initial velocity. then using kinematics (vf^2 = vi^2 + 2a<em>delta</em>x) you can solve for <em>delta</em>x since you know your vf is going to be 0 (that's when it reaches its maximum) and you get a final answer of 38.3 m</p>
<p>Bobobabob:
1. The bullet doesn't stay embedded; it hits the block, transfers part of its momentum to the block while it's travelling through the block, and emerges with a reduced velocity. The block has gained momentum from the bullet, and winds up with a non-zero velocity. </p>
<ol>
<li><p>I don't see how you got 27.4 m/s as a final velocity; even if we follow your implicit assumption that the bullet stayed embedded in the block, you should get the answer in post#11.</p></li>
<li><p>Kinematics equation? There is no force of gravity involved in this problem. Are you commenting on the problem in post#1, or the one in post#11?</p></li>
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<p>I think he was talking about the post#1, because that problem had the word "embedded" in it.</p>
<p>optimizerdad, you're so good with physics. Do you think you can help me with these 2 questions? These are confusing to me, because they require knowledge of multiple topics.</p>
<ol>
<li><p>A tennis ball of mass m = 0.060 kg and speed v = 25 m/s strikes a wall at a 45 degree angle and rebounds with the same speed at 45 degrees. What is the impulse given the wall?</p></li>
<li><p>A 115-kg fullback is running at 4.0 m/s to the east and is stopped in 0.75 s by a head-on-tackle by a tackler running due west. Calculate (a) the original momentum of the fullback, (b) the impulse exerted on the fullback, (c) the impulse exerted on the tackler, and (d) the average force exerted on the tackler.</p></li>
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<p>Hey - these are the same questions you'd posted in your other thread. Go read the responses there... :)</p>