1) An object has a kinetic energy of 262 J and a momentum of magnitude 25.4 kg ·
ID: 1461083 • Letter: 1
Question
1) An object has a kinetic energy of 262 J and a momentum of magnitude 25.4 kg · m/s. (a) Find the speed of the object. (b) Find the mass of the object.
2) A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor? (Magnitude and direction.)
3) A 2.90-kg steel ball strikes a massive wall at 10.0 m/s at an angle of = 60.0° with the plane of the wall. It bounces off the wall with the same speed and angle. If the ball is in contact with the wall for 0.220 s, what is the average force exerted by the wall on the ball? (Take to the right as the +x-direction and up as the +y-direction.) (Magnitude and direction).
Explanation / Answer
1)
ke=1/2mv^2
262=1/2mv^2
mv^2=524
m=524/v^2
momentum= p
p=mv
25.4=mv
now substitute in m
25.4=(524/v^2)v
25.4=524/v
25.4v= 524
v=524/25.4
speed of the object.
v =20.629921
now substitute in v into momentum equation
25.4=20.62m
mass of the object
m=1.23181
2)
down (-ve) & Up (+ve)
------- O PE = mgH (dropped from) u=0, KE=0
--- O PE = mgh v=0, KE=0
...... rising . ^ O v1 ......O hitting (v2)
==== floor
mgH + 0 = 0 + 0.5 mv1^2
v1^2 = 2gH = 2*1.25*9.8 = 24.5
v1 = - 4.95 m/s
+++++++++
0 + 0.5 mv2^2 = mgh + 0
v2^2 = 2 gh = 2*0.6*9.8 = 11.76
v2 = + 3.43 m/s
========
change in momentum = m[v2 - (-v1)] = 0.12[3.43+4.95]
delta p = 1.0056 N-s
Impulse = delta p = 1.0056 N-s
this is the impulse given to the ball by floor---
3)
Change of momentum = 2 . 2.90 . 10 . cos 30 = 50.22kgm/s
Force = Impulse / time = 228.6 N
This force is perpendicular to the wall . without having the diagram, I'm thinking it must be in the x direction. At zero degrees to the x axis.
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