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Initially
Finally
35.0m/s
25.0m/s
θ
= 55.0
o
7.50x10
4
kg
6.50x10
4
kg
0.150m/s
0.0223m/s
v
2i
= ?
m
1
m
1
m
2
m
2
m
2
= 125g
v
1i
= 30.0cm/s
Initially
v
1f
= 17.5cm/s
m
1
= 175g
m
2
= 125g
Finally
v
2f
= ?
v
2i
= 0
Jim Guinn’s PHYS2211 Assignment #10 Solutions
Due Monday, Mar. 1, 2010
1. What is the magnitude and direction of the impulse that would have to be applied to a
65.0gball that is moving initially with a velocity of 35.0m/s in the y direction to change its
velocity to 25.0m/s at an angle of 55.0
o
above the +xaxis.
(Hint:
keep in mind that impulse is
a vector and solve for the components separately, first.)
We know that the impulse is given by
J
=
∆
p
=
p
f

p
i
, or in terms of the components,
J
x
=
∆
p
x
= p
x,f
 p
x,i
, and J
y
=
∆
p
y
= p
y,f
 p
y,i
, we have
J
x
= m v
x,f
– m v
x,i
= (0.0650kg) (25.0m/s) cos55.0
o
– (0.0650kg) (0) = 0.932kg m/s, and
J
y
= m v
y,f
– m v
y,i
= (0.0650kg) (25.0m/s) sin55.0
o
– (0.0650kg) (35.0m/s) = 3.61kg m/s .
So
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This note was uploaded on 09/25/2011 for the course PHYSICS 22 taught by Professor Lomant during the Spring '11 term at Georgia Perimeter.
 Spring '11
 lomant

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