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Exam4Solutions

# Exam4Solutions - Physics 2130 Final Exam Solutions 3 1 B...

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Physics 2130 Final Exam Solutions 1. 3 B The figure shows a car moving to the left, and slowing down. Given the coordinate axes shown, the car’s acceleration ~a could be A) 2 m / s 2 ( - ˆ x ) B) 2 m / s 2 (+ˆ x ) C) 2 m / s 2 ( - ˆ y ) D) 2 m / s 2 (+ˆ y ) x y v 2. An elevator is moving upward, but slowing down, with an acceler- ation of ~a = 1 m / s 2 downward. A block with mass m = 3 kg sits on the floor of the elevator, and a triangle with mass 1 kg sits on top of the block. (a) 4 Draw and label the forces on the triangle and the block. 3kg 1kg mg N 1 N 2 N 1 mg (b) 2 Find the normal force on the triangle due to the block. Let downward be the positive direction. The net force on the block is mg - N 1 , so mg - N 1 = ma = N 1 = m ( g - a ) = (1 kg)(9 . 8 m / s 2 - 1 m / s 2 ) = 8 . 8 N (c) 2 Find the normal force on the block due to the floor. The net force on the block is mg + N 1 - N 2 , so mg + N 1 - N 2 = ma = N 2 = N 1 + m ( g - a ) = 8 . 8 N+(3 kg)(9 . 8 m / s 2 - 1 m / s 2 ) = 35 . 2 N v a = 1m/s 2 3kg 1kg Page 2

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3. A ball is launched from the ground into the air with a velocity of ~v 0 = 3ˆ x + 4ˆ y . We will ignore air resistance. (a) 3 B The velocity ~v top of the ball at the top of its arc is A) 0 B) x C) y D) x + 4ˆ y (b) 3 How high is the ball above the ground at the top of its arc? The initial energy is E i = KE + PE = 1 2 mv 2 i , and the final energy is E f = KE + PE = 1 2 mv 2 f + mgh . Energy is conserved, so 1 2 mv 2 i = 1 2 mv 2 f + mgh = 1 2 ( v 2 i - v 2 f ) = gh = h = v 2 i - v 2 f g = (5 m / s) 2 - (3 m / s) 2 9 . 8 m / s 2 = 1 . 6 m v 0 v top ? h x y 4. A 5 kg block moving at 5 m / s to the right collides with a 1 kg block moving at 2 m / s to the left. After the collision, the 5 kg block is moving at 3 m / s to the right. (a) 3 What is the velocity of the 1 kg block after the collision? Momentum is conserved. If positive momentum points to the right, then the initial momentum is p i = (5 kg)(5 m / s) + (1 kg)( - 2 m / s) = 23 kg m / s . The final momentum is 23 = (5 kg)(3 m / s) + (1 kg) v = v = 8 m / s (b) 3 C After the collision, the kinetic energy of the system has A) increased B) stayed the same C) decreased Initial kinetic energy is KE i = 1 2 (5 kg)(5 m / s) 2 + 1 2 (1 kg)(2 m / s) 2 = 64 . 5 J . Final kinetic energy is KE f = 1 2 (5 kg)(3 m / s) 2 + 1 2 (1 kg)(8 m / s) 2 = 54 . 5 J 5kg 5m/s 1kg 2m/s before 5kg 3m/s 1kg v after Page 3
5. A 2 kg block is moving along a floor with coefficient of kinetic fric- tion μ K = 0 . 2. It has an initial speed of 4 m / s. The block slows to a stop. (a) 3 Find the force of friction on the block. The normal force on the block due to the table is mg = (2 kg)(9 . 8 m / s 2 ) = 19 . 6 N . F K = μ K F N = (0 . 2)(19 . 6 N) = 3 . 9 N (b) 3 How far does the block slide before coming to a stop? The initial kinetic energy of the block is KE = 1 2 mv 2 = 1 2 (2 kg)(4 m / s) 2 = 16 J . The friction does negative work W = F Δ x on the block, and steals it all away by the time it stops. Thus F Δ x = 16 J = Δ x = 3 . 9 N 16 J = 0 . 24 m 2kg !

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