(a) Find the recoil velocity of a 70.0-kg ice hockey goalie, originally at rest, who catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. (b) How much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible. (See Figure 2 )

*Strategy*

Momentum is conserved because the net external force on the puck-goalie system is zero. We can thus use conservation of momentum to find the final velocity of the puck and goalie system. Note that the initial velocity of the goalie is zero and that the final velocity of the puck and goalie are the same. Once the final velocity is found, the kinetic energies can be calculated before and after the collision and compared as requested.

*Solution for (a)*

Momentum is conserved because the net external force on the puck-goalie system is zero.

Conservation of momentum is

p
1
+
p
2
=
p′
1
+
p′
2
p
1
+
p
2
=
p′
1
+
p′
2
size 12{p rSub { size 8{1} } +p rSub { size 8{2} } = { {p}} sup { ' } rSub { size 8{1} } + { {p}} sup { ' } rSub { size 8{2} } } {}

(1)or

m
1
v
1
+
m
2
v
2
=
m
1
v′
1
+
m
2
v′
2
.
m
1
v
1
+
m
2
v
2
=
m
1
v′
1
+
m
2
v′
2
.
size 12{m rSub { size 8{1} } v rSub { size 8{1} } +m rSub { size 8{2} } v rSub { size 8{2} } =m rSub { size 8{1} } { {v}} sup { ' } rSub { size 8{1} } +m rSub { size 8{2} } { {v}} sup { ' } rSub { size 8{2} } } {}

(2)Because the goalie is initially at rest, we know v2=0v2=0 size 12{v rSub { size 8{2} } =0} {}. Because the goalie catches the puck, the final velocities are equal, or
v′
1
=
v′
2
=v′
v′
1
=
v′
2
=v′ size 12{ { {v}} sup { ' } rSub { size 8{1} } = { {v}} sup { ' } rSub { size 8{2} } =v'} {}. Thus, the conservation of momentum equation simplifies to

m1v1=m1+m2v′.m1v1=m1+m2v′. size 12{m rSub { size 8{1} } v rSub { size 8{1} } = left (m rSub { size 8{1} } +m rSub { size 8{2} } right )v'} {}

(3)Solving for v′v′ size 12{v'} {} yields

v′=m1m1+m2v1.v′=m1m1+m2v1. size 12{v'= { {m rSub { size 8{1} } } over {m rSub { size 8{1} } +m rSub { size 8{2} } } } v rSub { size 8{1} } } {}

(4)Entering known values in this equation, we get

v′
=
0.150 kg
70.0 kg
+
0.150 kg
35
.0 m/s
=
7
.
48
×
10
−
2
m/s
.
v′
=
0.150 kg
70.0 kg
+
0.150 kg
35
.0 m/s
=
7
.
48
×
10
−
2
m/s
.
size 12{v'= left ( { {0 "." "150"`"kg"} over {"70" "." 0`"kg"+0 "." "150"`"kg"} } right ) left ("35" "." 0`"m/s" right )=7 "." "48" times "10" rSup { size 8{ - 2} } `"m/s" "." } {}

(5)*Discussion for (a)*

This recoil velocity is small and in the same direction as the puck’s original velocity, as we might expect.

*Solution for (b)*

Before the collision, the internal kinetic energy KEintKEint size 12{"KE" rSub { size 8{"int"} } } {} of the system is that of the hockey puck, because the goalie is initially at rest. Therefore, KEintKEint size 12{"KE" rSub { size 8{"int"} } } {} is initially

KE
int
=
1
2
mv
2
=
1
2
0
.
150 kg
35
.0 m/s
2
=
91
.
9 J
.
KE
int
=
1
2
mv
2
=
1
2
0
.
150 kg
35
.0 m/s
2
=
91
.
9 J
.

(6)After the collision, the internal kinetic energy is

KE′
int
=
1
2
m
+
M
v
2
=
1
2
70
.
15 kg
7
.
48
×
10
−
2
m/s
2
=
0.196 J.
KE′
int
=
1
2
m
+
M
v
2
=
1
2
70
.
15 kg
7
.
48
×
10
−
2
m/s
2
=
0.196 J.

(7)The change in internal kinetic energy is thus

KE′
int
−
KE
int
=
0.196 J
−
91.9 J
=
−
91.7 J
KE′
int
−
KE
int
=
0.196 J
−
91.9 J
=
−
91.7 J

(8)where the minus sign indicates that the energy was lost.

*Discussion for (b) *

Nearly all of the initial internal kinetic energy is lost in this perfectly inelastic collision. KEintKEint size 12{"KE" rSub { size 8{"int"} } } {} is mostly converted to thermal energy and sound.

During some collisions, the objects do not stick together and less of the internal kinetic energy is removed—such as happens in most automobile accidents. Alternatively, stored energy may be converted into internal kinetic energy during a collision. Figure 3 shows a one-dimensional example in which two carts on an air track collide, releasing potential energy from a compressed spring. Example 2 deals with data from such a collision.

Collisions are particularly important in sports and the sporting and leisure industry utilizes elastic and inelastic collisions. Let us look briefly at tennis. Recall that in a collision, it is momentum and not force that is important. So, a heavier tennis racquet will have the advantage over a lighter one. This conclusion also holds true for other sports—a lightweight bat (such as a softball bat) cannot hit a hardball very far.

The location of the impact of the tennis ball on the racquet is also important, as is the part of the stroke during which the impact occurs. A smooth motion results in the maximizing of the velocity of the ball after impact and reduces sports injuries such as tennis elbow. A tennis player tries to hit the ball on the “sweet spot” on the racquet, where the vibration and impact are minimized and the ball is able to be given more velocity. Sports science and technologies also use physics concepts such as momentum and rotational motion and vibrations.

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