Background: I\'m out playing near a frozen pond. On the pond is a block of ice (
ID: 778085 • Letter: B
Question
Background: I'm out playing near a frozen pond. On the pond is a block of ice (initially at rest) with mass 13.0 kg. I hit the ice block with a snowball of mass 2.05 kg, thrown at a speed of 5.25 m/s, and since its sticky snow, the snowball sticks to the block. Friction may be ignored. Part A: Let pi be the total initial momentum of the block-snowball system, just prior to the ball hitting the block. Find the magnitude of pi. Express your answer numerically. Part B: Find vf, the magnitude of the final velocity of the block-snowball system, after the collision. Express your answer numerically. Part C: What is the change K=KfinalKinitial in the block-snowball system's kinetic energy due to the collision? Express your answer numerically in joules. Secure https//session.masteringphysics.com/myct/îtemView?assignmentProblemID-93417206 C MP10-Spring 2018 Item 2 C2 of 11 Constants 1 Periodi Table Part A m out playing near a frozen pond On the pond is a block of ce (initially at rest) with mass 13 0 kg Ihit the ice block with a snowtball of mass 2 05 kg, thrown at a speed of 525 m/s and since its sticky snow, the snowball sticks to the block Friction may be ignored ystem, just prior to the ball htting the block Find the magnitude of p Let p be the total initial momentum of the block-snowball s Express your answer numerically View Available Hint's) PA kg-m/s Part B Find ty, the magnitude of the final velocity of the block-snowbalil system, after the collsion Express your answer numerically e View Available Hintis) PartC O Type here to searchExplanation / Answer
mass of block m1 = 13.0 kg
mass of snowball m2 = 2.05 kg
initial speed of block v1i = 0
initial speed of snowball v2i = 5.25 m/s
initial momentum Pi = m1*v1i + m2*v2f = Pi = (13*0) + (2.05*5.25) = 10.7 kg m/s
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after collision the snowball sticks to block and move together with speed vf
final momentum Pf = (m1 + m2)*vf
from momentum conservation
total momentum is conserved
momentum before collision = momentum after collision
Pi = Pf
10.7 = (13 + 2.05)*vf
vf = 0.711 m/s <<<-------------ANSWER
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part (C)
initial kinetic energy Kinitial = (1/2)*m1*v1i^2 + (1/2)*m2*v2i^2 = (1/2)*2.05*5.25^2 = 28.3 J
final kinetic energy Kfinal = (1/2)*(m1+m2)*vf^2 = (1/2)*(13+2.05)*0.711^2 = 3.8 J
change in kinetic energy dK = Kfinal - kinitial = -24.5 J
DONE please check the answer. any doubts post in comment box
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