PLEASE ANSWER ALL PARTS AND SHOW ALL WORK! You are sitting in a sled at rest on

Question

PLEASE ANSWER ALL PARTS AND SHOW ALL WORK!

You are sitting in a sled at rest on a pond covered with nice, thick, frictionless ice. Your own mass is 92 kg and the mass of the sled when empty is 13 kg.

a) From shore, someone throws a baseball of mass 0.19 kg to you, and you catch it; the horizontal component of the ball's velocity is 27.1 m/s. What wil be the sled's and your speed with respect to the surface of the pond after you catch the ball? Note: Your answer should be in centimeters per second.

b). Different situation: suppose you are in the sled and it's alreadt moving to the west, at a speed of 3.3 cm/s. From the southern shore, your friend again throws a baseball of a mass of 0.19 kg, which you catch as it's traveling northward with a horizontal velocity component of 27.1 m/s. In this case, what will be the sled's (and your) speed after catching the ball? Note: Your answer should be in centimeters per second.

c). Once again, suppose you start at rest on the sled out on the pond. Suddenly, you jump off of the sled, after which the empty sled is traveling at a speed of 4.9 m/s.(Note that is now in meters per second). What will be your speed on ice after jumping off ( Note: this will be the same as the horizontal component of your velocity just as you jump.)

d). That empty sled is still moving at 4.9 m/s. A parachutist of a mass of 93 kg suddenly drifts straight vertically down from a helicopter, landing on the sled. What will be the sled's speed (including the parachutist) after he lands?

Explanation / Answer

a)

mass of the system before catching the ball is

92+13 =105 kg

mass after catching:

0.19+105=105.19 kg

momentum before catching

po = 0.19*27.2 = 5.168 kgm/s

momentum after catching:

p=(105.19)v

From law of conservation of momentum, we have

105.19v= 5.168

v = 0.049 m/s

= 4.9 cm/s

b)

The momentum before catching:

P =((105*3.3)/100) - 5.168

= -1.703kgm/s

momentum after catching:

P = 105.19v

105.19v=1.703

v = 0.016m/s

= 1.6 cm/s

c)

momentum before jumping is 0

momentum after jumping is

93v+13*4.9

Feom the law of conservation of momentum, we have

93v+13*4.9 =0

Thus the speed is

v=13*4.9/93=0.6849 m/s

d)

The speed is calculated as follows

13*4.9= (13+93) v

v = 0.6 m/s

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.