block sliding down ramp? please help!!!!?

Question

a small block ofmass M is realeased from rest at the top of the curved frictionless ramp. The block slides down the ramp and is moving with a speed of 3.5vnaught when it collides with a larger block of mass 1.5M at rest at the bottom of teh incline. the larger block moves to the right at a speed of 2v naught immediately after the collision express answers in terms of the given quantities and fundamental constants
(A) determine height h of the ramp from which the small block isreleased
(b) determine the speed of the small block after the collision
(C) the larger block slides a distance D before coming to rest. Determine the value of teh coefficient of kinetic friction mew between the larger block and the surface on which it slides
(d) indicate if it is a elastic or inelastic collision.

so i tried it out and here are my answers if they are right let me know if not please help me
(A)h=.5v^2/g
(b) deltap=fdeltaT
(c) mew(ma)=friction
(D) inelastic

Answer 1

A) Using conservation of energy since the ramp is frictionless
.5*M*3.5^2*v^2=M*g*h
h=.5*3.5^2*v^2/g
You left off the 3.5 coefficient as stated in the problem

B) In a dynamic like this, the most fundamental quantity is momentum.
M*3.5*v=1.5*M*2*v+M*v2
where v2 is the speed of the small block after collision
The M divides out to leave
.5*v=v2

C) Since we know the speed of the larger block after the collision to be 2*v, and the distance D, we can use energy and work. The work done by the frictional force over the distance D (1.5*M*g*µ*D) will equal the kinetic energy of the larger block after collision (.5*1.5*M*4*v^2)

µ=2*v^2/(g*D)

D) It probably is inelastic, but let's check to see what energy was lost to ensure the assumption is correct

Starting energy
.5*M*3.5^2*v^2

Ending energy
.5*M*v^2/4 + .5*1.5*M*4*v^2

let's divide out .5*M*v^2

Starting
12.25

Ending
6.25

Yup, inelastic

j

Answer 2

I don't think you have the right answers.

(A) To determine this use conservation of energy. Potential energy is turned into Kinetic energy. At the top of the ramp, the potential energy is Mgh. At the bottom of the ramp, the small block's Kinetic energy is 1/2*Mv^2 but v = 3.5vnaught. So Kinetic energy is 1/2*M(3.5vnauhgt)^2.

So the energry conservation equation says Mgh = 1/2*M(3.5vnaught)^2

or h = ((3.5vnaught)^2)/(2g)


(B) I have no idea what your answer means. You should be able to calculate the speed of the small block using conservation of momentum. Before the collision the momemtum is only due to the motion of the small block. So the momentum is mv = M(3.5vnaught). After the collision the momentum is due to motion of both objects. They tell you the velocity of the big object is 2vnaught and you know its mass is 1.5M, so it's momentum is 1.5M*2vnaught. This is 3Mvnaught. If momentum is conserved, then the small mass must still be travelling with a velocity of .5vnaught.

(C) The friction force is assumed to be independent of velocity and equal to mu*Normal force. The normal force is the weight of the big block, which is mass times acceleration of gravity. so F = mu*1.5Mg. This produces a constant deceleration of the mass, so you should be able to calculate that deceleration (negative acceleration) and apply the equations for constant force to calculate the distance travelled.

(D) It is an inelastic collision because it turns out that energy is not conserved.

Answer 3

1. Use the law of energy conservation. (1/2mvi^2 = mgh). Vi = initial velocity. 2. Calculate the loss of kinetic energies by friction. Since it came to rest, the final kinetic energy and velocity must be zero. (1/2mvi^2 - 1/2mvf^2 = E). Vf = final velocity. 3. Calculate the deceleration. Then use it in the formula : umg = ma where u is the coefficient. Answers : 1. V = 7.75 m/s 2. Energy = 60 J 3. U = 0.33 I hope I'm helping :P

Answer 4

Let vnaught be written as Vo

(a) As initial velocity of smaller block is zero,

using 2as=v^2 - u^2

s=h
v =3.5Vo
a=g

h=(3.5Vo)(3.5Vo) / 2g

h=12.25Vo^2 /2g= [ 6.125 Vo^2 / g ]
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(b)Applying law of coservation of momentum,

total initial momentum=M3.5Vo=3.5MVo

( because the bigger block is at rest)

Suppose after collision, the velocity of smaller block is xVo

total final momentum=MxVo +(1.5)M2Vo=MxVo +3MVo

total final momentum = total initial momentum

MxVo +3MVo=3.5MVo

x=0.5

Speed of the small block after collision=0.5 Vo
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(c)To find retardation of bigger block,

we use 2as=v^2 -u^2 where v=zero, s=D,u=2Vo

retardation a= -(2Vo) (2Vo) / 2D= -2Vo^2 / D

retardation due to friction =mue*g

mue*g=2Vo^2 / D

Co efficient of kinetic friction=mue=2Vo^2 / Dg
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(D)initial kinetic energy of smaller block=6.125MVo^2

initial kinetic energy of bigger block=zero

total initial kinetic energy=6.125MVo^2

final kinetic energy of smaller block=0.125MVo^2

final kinetic energy of bigger block=3MVo^2

Total final kinetic energy=3.125MVo^2

Loss in kinetic energy=3MVo^2

Hence collision is inelastic