A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a w
ID: 1318786 • Letter: A
Question
A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 6.0 m/s in the opposite direction. If the object is in contact with the wall for 0.002 s, what is the magnitude of the average force on the object by the wall? ANS: 8.4 kN
2. A 4.0-kg particle is moving horizontally with a speed of 5.0 m/s when it strikes a vertical wall. The particle rebounds with a speed of 3.0 m/s. What is the magnitude of the impulse delivered to the particle? ANS: 32 N?s PLEASE SHOW WORK
Explanation / Answer
1.
m= 1.2kg time of action = 0.002 sec
Initial velocity = 8m/s
Final velocity = 6m/s (in opposite direction)
We know v = u + at
a = (v-u)/t a = (6-(-8))/0.002 = 7000m/s2
F = ma =1.2*7000N = 8.4kN
2.
m= 4kg
let time of action = t sec
Initial velocity = 5m/s
Final velocity = 3m/s (in opposite direction)
We know v = u + at
a = (v-u)/t a = (3-(-5))/t = 8/tm/s2
F = ma = 4*8/t N = 32/t N
Impulse J = F*t = (32/t)*t N.s = 32N.s
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