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This is from mcatquestionaday.com
I understand every part except where they calculated F = 10N from..I think I'm missing something really obvious here, but I'm not sure. Why didn't they use F = 36N?
Half Passage: A 0.1 kg bullet is fired horizontally and travels at a constant velocity of 120 m/s before embedding itself in a (stationary) 5.9 kg wooden block.
Question 2: Assuming the block remains upright upon impact and experiences 36 N of frictional resistance, how far will the block travel before coming to a stop after the bullet makes contact?
Explanation: The answer is (B). This is an expansion from the previous question. To solve, we must determine the initial, horizontal velocity of the block-with-bullet and its deceleration. We solved the initial velocity through the momentum equation:
(mass1*velocity1) = (mass2*velocity2).
(0.1)(120)= (0.1+5.9)(velocity2
velocity2 = 0.1*120/6.0
= 2.0 m/s
The deceleration can be solved for using F = ma, where F = 10N (where/how did they get this value?) and m = 6.0 kg, giving a = 6.0 m/s2. To determine the distance traveled, we can use the kinematics equation:
Vf2 = Vi2 + 2ad
0 = 22 + 2(-6)d
d = -(2)2/(2*-6)
d = 0.33 m
I understand every part except where they calculated F = 10N from..I think I'm missing something really obvious here, but I'm not sure. Why didn't they use F = 36N?
Half Passage: A 0.1 kg bullet is fired horizontally and travels at a constant velocity of 120 m/s before embedding itself in a (stationary) 5.9 kg wooden block.
Question 2: Assuming the block remains upright upon impact and experiences 36 N of frictional resistance, how far will the block travel before coming to a stop after the bullet makes contact?
Explanation: The answer is (B). This is an expansion from the previous question. To solve, we must determine the initial, horizontal velocity of the block-with-bullet and its deceleration. We solved the initial velocity through the momentum equation:
(mass1*velocity1) = (mass2*velocity2).
(0.1)(120)= (0.1+5.9)(velocity2
velocity2 = 0.1*120/6.0
= 2.0 m/s
The deceleration can be solved for using F = ma, where F = 10N (where/how did they get this value?) and m = 6.0 kg, giving a = 6.0 m/s2. To determine the distance traveled, we can use the kinematics equation:
Vf2 = Vi2 + 2ad
0 = 22 + 2(-6)d
d = -(2)2/(2*-6)
d = 0.33 m
