Physics problem? Elastic collision?

May 2021 4 36 Report

Physics problem? Elastic collision?

A steel ball of mass 0.600 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal, as shown in the figure. At the bottom of its path, the ball strikes a 4.60 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

I can't figure it out. HELP.

Science & Mathematics / Physics

Jánošík

Verified answer

@ peter: yo... what have you been smoking man ???

m₁ = mass of the ball = 0.600 kg

m₂ = mass of the block = 4.60 kg

L = length of the cord = 0.700 m

g = acceleration by gravity = 9.81 m/s²

The potential energy of the ball at its initial position is:

PE = m₁ g L

That is also the kinetic energy of the ball at its lowest point,

so its speed at that point is:

KE = PE

m₁ v² / 2 = m₁ g L

v = √( 2 g L )

v = 3.7059 m/s

The equations for elastic collision are:

(see http://en.wikipedia.org/wiki/Elastic_collision )

m₁×u₁ + m₂×u₂ = m₁×v₁ + m₂×v₂

and

v₁ - v₂ = u₂ - u₁

where

m₁ = mass of the ball = 0.600 kg

u₁ = speed of the ball just before the collision = v = 3.706 m/s

v₁ = speed of the ball just after the collision = ?

m₂ = mass of the block = 4.60 kg

u₂ = speed of the bloc just before the collision = 0 m/s

v₂ = speed of the bloc just after the collision = ?

So:

(0.6 kg)×(3.706 m/s) + (4.6 kg)×(0 m/s) = (0.6 kg)×v₁ + (4.6 kg)×v₂

(2.224 kgm/s) = (0.6 kg)×v₁ + (4.6 kg)×v₂ . . . . . . . . equation 1

v₁ - v₂ = (0 m/s) - (3.706 m/s)

v₂ = v₁ + (3.706 m/s) . . . . . . . . . . . . . . . . . . . . . . . equation 2

Substitute equation 2 into equation 1.

(2.224 kgm/s) = (0.6 kg)×v₁ + (4.6 kg)×(v₁ + (3.706 m/s))

(2.224 kgm/s) = (0.6 kg)×v₁ + (4.6 kg)×v₁ + (17.05 kgm/s)

(-14.82 kgm/s) = (5.2 kg)×v₁

v₁ = -2.85 m/s

Substitue that value into equation 2.

v₂ = (-2.85 m/s) + (3.706 m/s)

v₂ = 0.855 m/s

You now found the VELOCITY of the ball and block just after the collision.

The opposite direction of their motion is indicated by their opposite sign.

ANSWER:

the SPEED of the objects after the collision is

ball : 2.85 m/s

block : 0.855 m/s

Sorry however i cant provide an explanation for with values coz it is going to take too much time to calculate . I will most effective tell u the process . Us mgh = exchange in kinetic power of ball for two positions of the , horizontal and vertical . So u can get speed of ball simply earlier than placing . Now use conservation momentum . So now u will hav one eqn. With 2 unknwns . Now as collision is elastic complete vigour earlier than and after the collision dame . So ok.E. Before collision = okay.E. After collision [i.E. Ke of both the bodies]. So now u will jav amother eqn. With equal 2 unkwns as above . Resolve these 2 eqns. Amf u will get required ans.

xoan

sorry but i cant explain with values coz it will take too much time to calculate . i will only tell u the procedure . us mgh = change in kinetic energy of ball for 2 positions of the , horizontal and vertical . so u can get velocity of ball just before striking . now use conservation momentum . so now u will hav one eqn. with 2 unknwns . now as collision is elastic total energy before and after the collision dame . so k.e. before collision = k.e. after collision [i.e. ke of both the bodies]. so now u will jav amother eqn. with same 2 unkwns as above . solve these 2 eqns. amf u will get required ans.