A bullet with a mass of 0.01kg is fired horizontally into a block of wood hanging on a string.The bullet stick
822266
Conservation of energy:  KE=PE\n1/2*Mb*Vb^2= (Mb+Mw)*g*h\nVb^2=2*(Mb+Mw)*g*h/Mb\nVb^2=2*(0.01+2)*10*0.1/0.01\nVb^2=402\nVb=20.05 m/s
158383
You have to shoot it from a far enough distance not to make it go farther than the half meter the bullet would cause it to go.It would also depend on where you hit the tin can. That would have an effect on it's travel.
818268
The one held by someone that can hit the target.
1053008
This question has been answered pretty well by everyone, the only thing I want to add is contridictory to a statement by TJ N. \n\n\nthe caliber of a bullet is expressed in a fraction of an inch.  not in millimeters a .22 caliber bullet is 22/100ths of an inch in diameter, a .50 caliber bullet is 50/100ths of an inch (or if you reduced the fraction its a full 1/2 inch in diameter).  Sometimes ammunitions are interchangable with NATO ammuntions IE. Remington .223 = NATO 5.56 mm (its 223/1000ths of an inch measured metricly its 5.56 mm. \n\nthe problem with TJ Ns "years of shooting and hunting" experience is that a ".22 caliber bullet" - commonly a squirel/vermin hunting round would be 0.86 inch in diameter.... which wouldn't leave much squirel meat after you hit it with TJ's .22 caliber gun... Which leaves me with more questions about how much of a recoil his .22 caliber gun has? I've fired the .50 caliber Barret Sniper rifle before, as well as the M2 HB .50cal MG. (they both fire the same basic .50 caliber (50/100ths or 1/2 inch round) The M2 has to be mounted on some sort of mount (vehicle or sandbagged Tripod) to fire.  The Barret, I've heard of people firing it standing up.... but the only way I could get behind that thing was in the prone, and after about 15 rounds through that thing I had a bruise on my shoulder the size of Texas. I can't imagine what TJ's .22 caliber that's 86/100ths of an inch would do to me....\n\nnow the difference between the calibers... obviously is the size (diameter) of the bullet.  what's considered High caliber and low caliber?  I suppose thats a matter of opinion.  Some people say anything above .308 is large caliber, and I tend to agree with that but that doesn't really give an indication of damage caused by a bullet for example a .30 caliber (7.62 mm) fired from a M240B MG would tend to cause greater damage to soft tissue at long ranges(greater than 50 meters) than a .45 caliber fired from a M1911 handgun. \n\nAnd yes various rifles fire various caliber rounds.
802288
the impuls (m*v) is conserved, so the impulse in one direction;\n0.150*20*sin(30deg)\nEquals the impulse in the other direction\n50*v_recoil\nthe recoil velocity is thus 0.03 m/s
59102
If the bullet is fired in the horizontel direction, both object will reach ground at the same time.. because they are reaching ground due to vertical gravitational acceleration only which is same for both.\n\nBut if the bullet is fired in the downward direction and at an angle downwards, it will reach grounf first because it is having a initial velocity as well whereas the initial velocity for the other object is zero.. The reverse holds true if the bullet is fired upwards.
925506
You need to consider conservation of momentum and conservation of energy.\n\nConservation of Momentum says that the momentum before tha collision is equal to the momentum after the collision.\n\nm*v1 = (m + M)*v2\n\nwhere m is the mass of the bullet, M is the mass of the block, v1 is the speed of the bullet, and v2 is the speed of the bullet/block system. Therefore,\n\nv2 = m*v1/(m + M)\n\nTo find out how high the bullet/block system rises, consider conservation of energy. The kinetic energy of the bullet/block is converted to potential energy as the system rises so that\n\n0.5*(m + M)*v2^2 = (m + M)*g*h\n\nwhere g is the gravitation acceleration and h is the height to which the bullet/block rises. Solving for h\n\nh = v2^2/(2*g).
730340
Projectile range would be the equivalent of horizontal distance.\nThe forces involved are gravity (acting on the projectile's mass in the vertical direction), and the "push" from the initial thrust.\n\nAs the angle of the firing position increases above the horizontal, more of the initial push is used to combat gravity.\n\nIf the angle is too low, the projectile will meet the ground before it's maximum range (not enough action against gravity).\n\nFor most projectiles, the optimum angle is between 42 and 45 degrees (depending on the strength of the initial impetus). This is because the trajectory approximates a parabola.\n\nWith the given choices, the answer will have to be 45 degrees (answer B)...
735664
we know m=0.0043kg, Vfinal=304m/s, d=0.86m and Vinitial=0\n\nwhat we need is the value of acceleration ( remember F=ma )\n\nso the formula we need is\n\nvfina^2=viinitial^2 + 2ad\n304^2=0 + 2*0.86*a\na=(304^2)/(2*0.86)\na=53730.23256\n\nso F=m*a=53730.23256*.0043=231.04 N\n\nForce is 231.04 N\n\n*****************************************************\n\nHere's an alterative if you want to double check  \n\nthe kinetic energy (KE)=1/2 *m*(velocity)^2\n\nthe KE of the bullet must equal the Work needed to fire it\n\nKE=0.5*0.0043*304^2 =198.6944 j\n\nWork=F*d=198.6944\n\nso F=(198.6944)/0.86=231.04 N\n\nAgain we get  force = 231.04 N\n\nSo we're pretty sure this is the correct answer
293083
45 degrees is the optimal angle to launch a projectile for maximum horizontal displacement.
453732
You wouldn't hit the snake, and the water has nothing to do with the trajectory of the bullet.\nWhat you want to keep in mind is the refraction of light in air compared to water.\nFor example, if you put a stick half in and half out of water, you'll notice that it doesn't look straight anymore.  The part in the water will be offset from the part in the air.\nTherefore if you aim directly at the snake, you'll miss.
501416
The bullet is packed with gunpowder. When the firing pin inside the pistol strikes the bullet, the gunpowder ignites and propells the bullet. The mass of the pistol is irrelevant, because the gunpowder inside the bullet is providing the force to push the bullet.
753733
You put a bullet in a gun and set it on a table and leave it. I will guarantee that gun will never kill no one. You put that same gun in the wrong person's hands and it will kill someone. So therefore it must have something to do with the person.
164638
I must bebunk the answerers who think the bullet will be travelling faster without the absence of gravity - the answer is yes and no - yes if compared to a bullet fired vertically on earth or at any angle, no if compared to bullet fired horizontal to the ground on earth.\n\nIf bullet is shot horizontally here on earth, and then shot in space, both bullets will travel at the same speed. The bullet travelling horizontally on earth will have gravity only affect its vertical movement, hence it will be pulled faster and faster towards the ground, but its horizontal velocity will stay the same.\n\nStraightdope has covered this in an article:
411752
One possibility would be frangible ammo, which won't penetrate through walls and is non-toxic (no lead).  I don't know if you can get rubber bullets anywhere, but that may be a bad idea (it would just make the intruder madder).\n\nhttp://www.precisionammo.com/\nhttp://accutecusa.com/\nhttp://www.precisioncartridge.com/frangible.html\nhttp://www.iccammo.com/products.html
834149
The period T is 0.9s\nT = 1/f = 2pi/w so w = 2pi/T\n\nAs the movement is circular, the total of the forces is in this plan and to the centre and the value is mw²R.\n\nSo P + S = mw²R where P is the weight and S the tension of the string and R the radius and a the angle between the string and the vertical\nIn projection, we have :\nmw²R = S sin a\nmg = S cos a\n\nDividing the 2 equations :\ntg a = w²R/g\nand R = L sin a\n\nso w² = g/(Lcos a)\nso cos a = g/Lw² = g T²/(L 4 pi²) \n\nso cos a= 9.8 0.9²/ (0.3 4 3.14²) = 0.6702\n so a = 47.91°  or a =0.83 rad\n\nBut this angle is the angle between the string and the vertical so angle between the string and the horizontal is 90-a = 42.09°
757988
a) \nKE if the block and bullet= 0.5(m+M)u²\nuse hookes law to find energy stored in spring = 0.5kd²\nthese are equal due to E conservation between block and spring, so\nu=sqrt[kd²/(m+M)]\n\nmultiply this my the mass of the block and bullet to find this couple's momentum, u × (m+M).\n\nthis answer is equal to the momentum of the bullet, since momentum is conserved for perfectly inelastic collision. to find bullet speed, v= u(m+M)÷m. remember to sub the derived terms into u.\n\nb) sub the values (convert to SI units first) into the final equation. the equation is unecessarily long because of the spring, which introduces additional variables. in practice, a pendulum is used.\n\nc) using the v value found in a, calculate the KE of the bullet using 0.5mv², then compare to the KE of the bullet and block combined. the latter will be smaller because energy is lost and heat/sound during collision.