Problem:

A bit of mass m1 and also velocity u1 collides with a particle of massive m2 at rest. The two particles rod together. What portion of the original kinetic power is lost in the collision? (Simplify the answer as much as possible.)

Solution:

Concepts:Momentum conservationReasoning:In one inelastic collision kinetic power is not conserved, but momentum is conserved.Details the the calculation:m1u1 = (m1 + m2)v.Ef = ½ (m1 + m2)v2, Ei = ½ m1u12.Fraction of energy lost = (Ei - Ef)/Ei = 1 - m1/(m1+ m2) = m2/(m1 + m2).Problem:

A bit of fixed m traveling with (non-relativistic) velocity u1 makes a head-on collision with a second particle of massive M, i m sorry is at rest in the laboratory. If the collision is totally inelastic, what portion of the original kinetic power remains after ~ the collision?

Solution:

Concepts:Inelastic collisions, momentum conservationReasoning:In inelastic collisions: momentum is conserved, mechanical power is not conserved.Details of the calculation:Assume the particle is at first traveling in the positive x-direction.Before the collision: pi = mu1, Ei = ½mu12.After the collision: pf = (m+M)v = mu1. Ef = ½(M+m)v2 = ½u12.Ef/Ei = m/(M+m).Problem:

A 36 g bullet through a speed of 350 m/s strikes a 8 cm thickfence post. The cartridge is retarded by one average force of 3.6*103 Nwhile traveling every the way through the board.(a)What rate does the bullet have actually when it emerges?(b)How many such boards could the cartridge penetrate?

Solution:

Concepts:Newton"s 2nd law, F = ma, activity with consistent accelerationin one dimensionReasoning:For activity with constant acceleration the equations the kinematicsyieldvxf2 = vxi2 + 2ax(xf- xi).Details the the calculation:(a) (xf - xi) = 0.08 m, vxf2 =(350 m/s)2 - 2*(0.08 m)(3.6*103 N)/( 36*10-3kg),vxf = 326.34 m/s.(b) vxf = 0, (350 m/s)2 =2*d(3.6*103 N)/( 36*10-3 kg), d = 0.6125 m.# of board the bullet deserve to penetrate = 7, (d/0.08 = 7.65)orWork-kinetic energy theorem: d = ½mvi2.Problem:

Board A is put on plank B together shown. Both boards slide, without relocating with respect to each other, along a frictionless horizontal surface at a speed v. Board B access time a relaxing board C "head-on." after the collision, plank B and also C move together, and board A slides on height of board C and also stops its activity relative come C in the position shown on the diagram. What is the length of every board? All 3 boards have the exact same mass, size, and shape. That is recognized there is no friction between boards A and B; the coefficient ofkinetic friction between boards A and also C is μk.

Solution:

Concepts:Inelastic collisions, momentum conservation, frictionReasoning:Board B and C collide inelastically. Plank A slides over boards B and also C, and also we have sliding friction. Solve the problem in a inertial frame
.Details of the calculation:After the collision plank B and also C stick together and also move with rate v/2 in the lab structure (momentum conservation). Over there is no friction in between A and also B. Board A moves with rate v in the laboratory frame. The CM structure moves v velocity 2v/3 in the direction of the right. In the CM frame board A moves v velocity v/3 towards the right and boards B and also C move with velocity v/6 towards the left right after the inelastic collision. Permit us work-related in the frame. Let the best edge of board A be at x" = 0 at t = 0, appropriate after the collision. At time t, as soon as the right edge of the board A is in ~ x", the force on the plank is the force of friction μN = μmg3x"/(2L) directed towards the left.L is the length of every board, 3x"/(2L) to represent the portion of the height board the is sustained by the surface ar with friction.The acceleration is a = F/m, d2x"/dt2 = -μg3x"/(2L).The systems to this differential equation is x" = x"0sinωt, ω = (3μg /(2L))½, v" = ωx"0cosωt,with the initial conditionsx"(0) = 0, v"(0) = v/3 = ωx"0.x"0 = v/(3ω) = v"(0)/(3μg/(2L))½.When x" = x"0, v" = 0. X"0 therefore is the (2/3) of length of the board, x"0 = 2L/3.2L/3 = v"(0)/(3μg/(2L))½, 2L2/9 = v"(0)2L/(3μg).The length of the board is L = 3v"(0)2/(2μg) = v2/(6μg)You can additionally use the work-kinetic energy theorem. We have in the centimeter frameF = -Cx", C = 3mμg/(2L). C(2L/3)2/2 = ½m(v/3)2, together = v2/(6μg).Problem:

An hour glass sit on a scale. At first all the sand (mass m) in the glass (mass M) is organized in the upper reservoir. At t = 0, the sand is released. If it exit the top reservoir in ~ a constant rate dm/dt = -λ, attract (and label quantitatively) a graph showing the analysis of the scale at all times t > 0.

You are watching: Kinetic energy is the force that needs to be dissipated in a collision

Solution:

Concepts:Newton"s second law, F
= dp/dtReasoning:The analysis of the range is determined by the amount of sand no falling freely and the pressure required to reduce the inert of an quantity of sand dmreaching the bottom to zero in the a time expression dt.Details the the calculation:Each grain of sand falls a street h. It as such falls for a time interval t1 = (2h/g)½. The very first grain hits the bottom at t = t1. For 0 1, the range reads W1 = (M + m)g - λgt.W1 to reduce linearly native (M + m)g come (M + m)g - λ(2gh)½.At time t2 = m/λ, all the sand has left the top reservoir. Because that t1 2, W2 = (M + m)g - λgt1 + v|dm/dt|.(M + m)g - λgt1 is the weight of the sand that is no falling and v|dm/dt| is the force required to mitigate the momentum of the amount of sand |dm| = λdt getting to the bottom in the time interval dt with speed v = (2gh)½ to zero in the time interval dt.v|dm/dt| = (2gh)½λ. λgt1 = λg(2h/g)½ = (2gh)½λ.Therefore W2 = (M + m)g. W2 is constant.At time t3 = t2 + t1 all the sand has actually reached the bottom. For t2 3, W3 = W2 + λg(t - t2) .W3 rises linearly from (M + m)g to (M + m)g + λ(2gh)½.For t > t3, W4 = (M + m)g = constant.
Problem:

A 15.2 g bullet hits a 0.463 kg block from below. The initial speed of the bullet is 624 m/s and also it emerges from the block in ~ 131 m/s. (a) how high go the block rise?(b) If the block is 2.34 centimeter thick, estimate the average force on the block. Assume the the bullet passes totally through before the block moves appreciably.

Solution:

Concepts:Momentum conservation, impulseReasoning:In the collision in between the bullet and also the block inert is conserved. The collision time is very short.This is a 1-dimensional problem.Details of the calculation:(a) m = 0.0152 kg , M = 0.463 kg, mvi = mvf + Mv.v = m(vi - vf)/M = 0.0152*(489 m/s)/ 0.463 = 16 m/s.The initial speed of the block is v = 16 m/s.It rises to a height h = v2/(2g) = 13.15 m.(b) The block obtain an advertise Δp = Mv in the time interval Δt.Δt is the moment it bring away the cartridge to pass with the block, Δt = 2*(0.0234 m)/(vi + vf) = 6.23*10-5 s, presume uniform deceleration.F = Δp/Δt = 1.19*105 N.

Collisions in 2D

Problem:

A particle of massive m strikes a frictionless, smooth, hard surface at an edge θ indigenous the surface ar normal. The bit bounces off through coefficient of restitution ε. (The coefficient of restitution is characterized as the ratio of the magnitudes of the velocity components along the normal to the plane of contact after and also before the collision.) uncover the fag angle for the fragment as it leaves the surface.

Solution:

Concepts:Definition: coefficient the restitutionReasoning:This is an inelastic collision. We space asked to uncover the rebound angle when provided the coefficient that restitution.Details the the calculation:ε = |v2|/|v1|, v1 is the velocity component of the fragment toward the floor prior to collision and v2 is the velocity ingredient of the particle away native the floor after ~ the collision.Before the collision: tanθ = |vx|/|vy|.After the collision: tanθ" = |vx"|/|vy"| = |vx|/(ε|vy|) = (1/ε)tanθ.Since ε Problem:

A 90 kg fullback running eastern with a speed of 5 m/s is tackled by a 95 kg foe running north through a speed of 3 m/s. If the collision is perfect inelastic, calculation the speed and the direction that the players just after the tackle.

Solution:

Concepts:Inelastic collision, conservation of momentumReasoning:If a component of the total external force acting top top a system is zero, then the matching component that the complete momentum is conserved. Us assume the the horizontal component of the outside force deserve to be neglected during the collision.Details that the calculation:The initial momentum of player 1 is p
1= (90 kg 5 m/s)i = (450 kgm/s)i.The initial inert of player 2 is p2 = (95 kg 3 m/s)j = (285 kgm/s)j.Momentum is conserved, the last momentum p of both football player is p = p1 + p2.p = (m1 + m2)v.v = (2.432 m/s)i + (1.54 m/s)j.v2 = 8.29(m/s)2, v = 2.88 m/s.The rate of the football player after the collision is 2.88 m/s.tanθ = py/px = 285/450 = 0.63, θ = 32.34o.Their direction the travel makes an edge θ= 32.34o through the x-axis. (The x-axis is pointing east.)Problem:

After a totally inelastic collision between two objects of same mass, each having initial rate v, the two relocate off along with speed v/3. What was the angle in between their early directions?

Solution:

Concepts:Conservation that momentumReasoning:The outside force exhilaration on the device is zero, the total momentum is conserved.Details of the calculation:
p1x + p2x = pfx = pf, p1y+ p2y = 0.2mv cosθ = 2mv/3, cosθ = (1/3), θ = 70.5o.The angle between their initial direction is 2θ = 141o.Problem:

The mass of the blue puck is 20% better than the fixed of the eco-friendly one. Prior to colliding, the pucks strategy each other with equal and also opposite momenta, and also the green puck has actually an initial rate of 10 m/s. Uncover the speed of the pucks ~ the collision, if half the kinetic energy is lost during the collision.

Solution:

Concepts:Inelastic collisions, inert conservationReasoning:The collision is inelastic, due to the fact that energy is not conserved.The full momentum that the 2 pucks is zero prior to the collision and also after the collision.Let bit 1 be the green puck and also particle 2 it is in the blue puck. Before and after the collision the ratio of the speeds is v2/v1= m1/m2 = 1/1.2.Details of the calculation:The last kinetic energy of the system equates to ½ time its initial kinetic energy.½m1v1i2+ ½m2v2i2 = 2(½m1v1f2+ ½m2v2f2).m1v1i2+ m2v2i2 = 2(m1v1f2+m2v2f2).m1v1i2 + 1.2m1(v1i/1.2)2= 2(m1v1f2 + 1.2m1(v1f/1.2)2).m1(v1i2 + v1i2/1.2) = 2m1(v1f2 + v1f2/1.2).v1i2 = 2v1f2, v1f = 0.707v1i.v1f = 7.07 m/s, v2f = 5.89 m/s.Problem:

A wooden block of mass M hangs on a massless rope of size L. A cartridge of fixed m collides through the block and remains within the block. Discover the minimum velocity of the cartridge so that the block completes a full circle about the suggest of suspension.

Solution:

Concepts:Inelastic collisions, preservation of momentum, circular motionReasoning:Let united state treat the block/bullet mix as a suggest particle. The block and also bullet will certainly start relocating in a circle. The forces on the block/bullet room the force of gravity and the anxiety in the rope.If the initial rate is big enough so that the block/bullet benefit a height greater than L, climate the centripetal acceleration essential to keep the block/bullet relocating on the one is partly provided by gravity and also the stress decreases. Whenthe tension vanishes then the block/bullet behaves like a totally free particle in a gravitational ar executes projectile motion.Details the the calculation:Conservation the momentum:mvbullet = (M + m)v, v = mvbullet/(M + m)When the block/bullet reaches the top of the circle through a speed vtop, so that the centripetal acceleration vtop2/L > g, then it will completes a full circle about the suggest of suspension. ½(M + m)v2 - 2(M + m)gL = ½(M + m)vtop2 vtop2 = v2 - 4gL = 2(vbullet)2 - 4gL.We need 2(vbullet)2/L - 4g > g --> vbullet > (5gL)½ (M + m)/m.Problem:

A car of total mass M1 = M and velocity v1 provides a completely inelastic collision at time t = 0 through a second car of massive M2= 2M at rest. Prior to the collision a point object of mass m 1 will certainly the tiny mass lose contact with the surface ar of the cavity?

Solution:

Concepts:Inelastic collisions, Newton"s laws, preservation of momentum, circular motion, frame transformationsReasoning:Before the collision auto 1 and also mass m move with uniform velocity v1. The advertise of the collision alters the velocity of vehicle 1, and after the collision auto 1 moves with uniform velocity v2. The small mass moves with velocity v = v1 - v2 v respect to vehicle 1 automatically after the collision. ~ the collision the rest framework of auto 1 is an inertial frame and also we deserve to analyze the trouble in that frame.Details of the calculation:Conservation that momentum: Mv1 = 3Mv2, v2 = v1/3, v = v1 - v2 = 2v1/3.In the rest frame of auto 1 instantly after the collision we have actually a particle of massive m at the bottom of a cavity of radius r relocating with velocity v. The fragment will start moving in a circle. The pressures on the fragment are the pressure of gravity and the normal force from the wall of the cavity. If the initial rate is huge enough so the the bit gains a height greater than r, then the centripetal acceleration essential to keep the particle moving on the one is partly listed by gravity and also the normal pressure decreases. If normal force vanishes climate the bit behaves favor a free particle in a gravitational field and looses call with the wall.

When ½mv2 ½, or v1 ½ the bit will no break call with the wall.When the bit reaches the top of the cavity through a rate vtop, so the the centripetal acceleration vtop2/r > g, climate it will not break call with the wall. Then ½mv2 = 2mgr + ½mvtop2 > 2mgr + ½mgr = (5/2)mgr.v > (5gr)½, v1 > (3/2)(5gr)½.For initial velocities (3/2)(2gr)½ 1 ½ the bit breaks call with the wall.

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Problem:

A brick is thrown (from soil level) with speed V at an edge θ v respect to the (horizontal) ground. Assume the the long confront of the brick continues to be parallel come the soil at all times, and also that over there is no deformation in the floor or the brick once the brick hits the ground. If the coefficient the friction in between the brick and also the floor is μ what need to θ be so the the brick travel the maximum full horizontal distance prior to finally comes to rest?

Solution:

Concepts:Projectile motion, impulsive forcesReasoning:The normal force from the ground has to carry out an advertise to mitigate the upright component of the momentum of the brick to zero.Details of the calculation:Let V it is in the early stage speed. The horizontal speed and initial vertical speed space then Vcosθ and also Vsinθ, respectively. The horizontal distance traveled in the waiting is the standarddair = 2V2sinθcosθ/g.To discover the street traveled along the ground, us must recognize the horizontal speed just after the affect has occurred. The normal force, N, native the ground is what to reduce the vertical rate from V sinθ to zero, throughout the impact. So we have ∫ N dt = m V sinθ,where the integral runs over the time of the impact. But this normal force (when multiplied by μ to provide the horizontal friction force) additionally produces a suddenly decrease in the horizontal speed, during the time the the impact. For this reason we have actually m ∆vx = -∫ μN dt = -μ m Vsinθ,which suggests ∆vx = -μ Vsinθ.(We have actually neglected the effect of the mg gravitational force during the brief time that the impact, since it is lot smaller 보다 the N impulsive force.) Therefore, the brick begins its sliding activity with speedv = Vcosθ - μ Vsinθ.(Note the this is true only if tanθ The friction force from this point on is μmg, therefore the acceleration is a = -μg. The distance traveled along the ground isdground = (Vcosθ - μ Vsinθ)2/(2μg).We want to uncover the angle the maximizes the full distance, dtotal = dair + dground.d(dtotal)/dθ = 0. D(sinθcosθ + cos2θ/(2μ) + μ sin2θ/2)/dθ = 0.cos2θ - sin2θ + sinθ cosθ (μ - 1/μ) = 0. Cos(2θ) + sin(2θ)½(μ - 1/μ) = 0.tan(2θ) = 2μ/(1 - μ2). Utilizing the trigonometric identity, tan(2θ) = 2 tanθ/(1 - tan2θ) we have actually tanθ = μ.Taking the derivative v respect come θ, we watch that the maximum total distance is achieved when tanθ = μ. Note, however, that the over analysis is valid only if tanθ 1, then the optimal angle is θ = 45o. (The brick stops after the impact, and also μ = 45o provides the maximum worth for the dair.Problem:

A marble bounces under stairs in a continuous manner, hitting each step at the same place and bouncing the same height above each step.

The stair height equates to its depth (tread = rise) and the coefficient of restitution ε is given. Discover the crucial horizontal velocity and bounce height. (The coefficient the restitution is identified as ε = -vf/vi, where vf and also vi space the upright velocities just after and also before the bounce. Respectively).