As shown in the figure below, object m 1 = 1.75 kg starts at an initial height h
ID: 1346579 • Letter: A
Question
As shown in the figure below, object m1 = 1.75 kg starts at an initial height h1i = 0.345 m and speed v1i = 4.00 m/s, swings downward and strikes (in an elastic collision) object m2 = 4.50 kg which is initially at rest. Determine the following.
(a) speed of m1 just before the collision.
____ m/s
(b) velocity (magnitude and direction) of each ball just after the collision (Assume the positive direction is toward the right. Indicate the direction with the sign of your answer.)
____ m/s (m1)
____ m/s (m2)
(c) height to which each ball swings after the collision (ignoring air resistance)
____ m (m1)
____ m (m2)
Explanation / Answer
Solution: Given
Mass of the first ball m1 = 1.75 kg,
Initial height of the first ball h1i = 0.345 m,
Initial velocity of the first ball v1i = 4.00 m/s
Mass of the second ball m2 = 4.50 kg,
Initial height of the second ball h2i = 0.00 m,
Initial velocity of the second ball v2i = 0.00 m/s
Part (a)
Consider the lowest level of the above system as the our reference level where gravitational potential energy of the balls is zero.
Thus At the height h1i, the total energy of the ball 1 is,
T1i = (1/2)m1*v1i2 + m1*g*h1i
T1i = (1/2)(1.75kg)*(4.00m/s)2 + (1.75kg)*(9.81m/s2)*(0.345m)
T1i = 19.9228 J
Just before the balls strike, the gravitational potential energy of the m1 is converted into kinetic energy, thus final total energy of the m1 is as h1f = 0 m from our reference level.
T1f = (1/2)m1*v1f2 + m1*g*h1f
T1f = (1/2)m1*v1f2 + m1*g*(0m)
T1f = (1/2)m1*v1f2
Since total mechanical energy should be conserved, we have,
T1i = T1f
19.9228 J = (1/2)(1.75kg)*v1f2
v1f2 = 22.7689
v1f = 4.7717 m/s
Thus speed of m1 just before collision is 4.77 m/s
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Part (b)
(we will take the speed of the ball 1 to be v1i instead of v1f as of above for the sake of simplicity and convenient use of notation. Thus initial velocity of the m1 just before the collision is v1i = 4.7717 m/s)
Ball 2 is at rest just before the collision, v2i = 0.00 m/s
We want to find the final velocities of the ball 1 and 2 just after the collision take place.
Since the collision is elastic in nature, then both the linear momentum and kinetic energy are conserved in above case.
m1*v1i + m2*v2i = m1*v1f + m2*v2f ---------------------------------------(1)
(above equation is for conservation of linear momentum)
(1/2)*m1*v1i2 +(1/2)*m2*v2i2 = (1/2)*m1*v1f2 + (1/2)*m2*v2f2 ------------------(2)
Above statement is for the conservation of total mechanical energy.
Noting v2i = 0.00 m/s; equation (1) and (2) can be solved for v1f = final velocity of the ball 1 and v2f = final velocity of the ball 2 just after the collision.
v1f = v1i*(m1 - m2)/(m1 + m2)
v1f = (4.7717m/s)*(1.75kg – 4.50kg) /(1.75kg + 4.50kg)
v1f = -2.0995 m/s
and
v2f = v1i*(2*m1)/(m1 + m2)
v2f = (4.7717m/s)*(2*1.75kg)/( 1.75kg + 4.50kg)
v2f = 2.6722 m/s
Thus ball 1 moves towards left with 2.0995 m/s and ball 2 moves towards right with 2.6772 m/s.
Thus the answers with signs are
v1f = -2.0995 m/s for m1
v2f = 2.6722 m/s for m2
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Part (c)
After collision, the both the masses will rise to a height where there kinetic energy is converted into potential energy,
For mass 1
(1/2)*m1*v1f2 = m1*g*h1f
h1f = v1f2/(2*g)
h1f = (-2.0995m/s)2/(2*(9.81m/s2))
h1f = 0.2247 m
Thus mass 1 will rise to a height of 0.2247 m after collision.
For mass 2
(1/2)*m2*v2f2 = m2*g*h2f
h2f = v2f2/(2*g)
h1f = (2.6722m/s)2/(2*(9.81m/s2))
h1f = 0.3639 m
Thus mass 2 will rise to a height of 0.3639 m after collision.
Thus the answers are,
0.2247 m for m1
and
0.3639 m for m2
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