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Olaf is standing on a sheet of ice that covers the football stadium parking lot in...

Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible friction between his feet and the ice. A friend throws Olaf a ball of mass 0.400 kg that is traveling horizontally at 11.3 m/s. Olaf's mass is 75.0 kg.

(a) If Olaf catches the ball, with what speed v_f do Olaf and the ball move afterward?

(b) If the ball hits Olaf and bounces off his chest horizontally at 8.90 m/s in the opposite direction, what is his speed v_f after the collision?
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Answer #1
Concepts and reason

This problem is based on the concept of conservation of linear momentum.

Since there is no external force is action on the system therefore total momentum of the system remain conserved. First equate the total final and initial momentum of the system and then calculate final velocity of the both Olaf and the ball.

Fundamentals

Momentum: momentum is a vector quantity which is equal to the product of mass and velocity of the body. This can be written as follows;

P=mv

Here,is the momentum,is the velocity of the mass and is the mass of the body.

Law of conservation of momentum states that total momentum of the system remains conserved if there is no external force acting on the system. It is represented by the expression as follows:

P: = Pc

Here, is initial momentum and is final momentum.

(a)

Calculate the velocity of ball and Olaf as follows;

The initial momentum of ball and Olaf can be written as follows;

Pu=mu, + Muil

Here, is the initial momentum of ball and Olaf, is the mass of the ball,is the initial velocity of the ball,is mass of the Olaf, and is the intimal velocity of the Olaf.

Substitute 11.3 m/s
for, 0.400 kg
for , and 0 forin the equation of.

Pa =(0.400 kg)(11.3 m/s)+M(0)
= 4.52 kg.m-s

The final momentum of the ball and Olaf can be written as follows;

Pri = mv. + Mv

Here, is the final momentum of the ball and Olaf,is the mass of the ball, is the final velocity of the ball,is mass of the Olaf and is the final velocity of the Olaf.

Substitute for , for , 0.400 kg
for , and75.0 kg
forin the equation of.

Pa =(0.400 kg)(v)+(75.0 kg)(v)
=v(75.400 kg)

From the law of conservation of momentum the initial and final momentum of the system should be equal. This can be written as follows;

Pri =P

Here, is final momentum of the system and is the initial momentum of the system

Substitute (75.400 kg)
forand4.52 kg.m-s
for and rearrange the equation for.

v (75.400 kg) = 4.52 kg.m.s
4.52 kg.m.s
75.400 kg
= 0.0600 m/s
V=

Hence final velocity of the Olaf and ball is after the collision is0.0600 m/s
.

(b)

Law of conservation of momentum is represented by the expression as follows;

P: = Pc

Substitute my, +Mv.
forand mv + Mvi
for.

mv, + Mv, = mv + Mv.

Here, is mass of ball, is mass of Olaf, is velocity of ball before collision, is velocity of Olaf before collision, is velocity of ball after collision and is velocity of Olaf after collision.

Substitute11.3 m/s
for,0 for , -8.90 m/s
for, for , andforin equation mv, + Mv, = mv + Mv.
and solve for .

(0.400 kg)(11.3 m/s?) +0 = (0.400 kg)(-8.90 m/s²)+(75.0 kg)v.
v=0.108 ms

Therefore, velocity of the Olaf is 0.108 m.s?

Ans: Part a

The Speed of ball and Olaf is 0.0600 m/s

Part b

The Speed of Olaf is0.108 m.s?
.

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