2624 Words11 Pages

Physics 221 Summer 2012
HOMEWORK #3
Due Monday July 2, 2012
1
James Bond (90 kg), outﬁtted with perfectly matching skis and skiware, is at the top of a steep slope that a secret spy like him can easily handle. He lets himself go from rest and smoothly slides down the h = 15 m high hill. A big parking lot lies at the bottom of the hill. Since the parking area has been cleared of snow, the friction between the ground and the skis brings our hero to a halt at point D, located at a distance d = 12 m from point C. The descent can be considered frictionless. Take the potential energy to be zero at the bottom of the slope. (a) What is the mechanical energy of James Bond at points A and D? (b) Determine the speed of Bond at position B*…show more content…*

A bullet of mass 10.0 g is ﬁred into the block and embeds in the block. The block rises a height of h = 5.00 cm. (a) Draw a ﬁgure showing the initial and ﬁnal positions of the block. (b) What was the speed of the bullet? (c) How much thermal energy is generated in the block? (d) If the bullet stops in a distance of 4.00 cm inside the block, estimate the force of friction. SOLUTION : (b) The problem consists of two stages. The ﬁrst stage is from before to after the impact. The second stage is from just after the impact to the ﬁnal position. In the ﬁrst stage, the linear momentum is conserved. In the second stage, the energy is conserved. Let us denote the initial speed of the bullet as v0 . So: pbefore = pafter mv0 = (m + M ) v (4) M +m v0 m M y=0 Then using the conservation of energy of the combined mass rising to height h after the impact: 1 (M + m) v 2 = (M + m) gh 2 ⇒ v = 2gh (a) L L y=h (5) Substituting v from eq. 5 into eq. 4 and solving for v0 : v0 = m+M m 2gh = 2.00 kg + 0.0100 kg 0.0100 kg 2 9.80 m/s2 (0.0500 m) = 199 m/s (c) Assuming no sound generated in the impact, all the energy lost will turn into heat energy. The heat energy generated is equal to the diﬀerence between the energy before the impact

A bullet of mass 10.0 g is ﬁred into the block and embeds in the block. The block rises a height of h = 5.00 cm. (a) Draw a ﬁgure showing the initial and ﬁnal positions of the block. (b) What was the speed of the bullet? (c) How much thermal energy is generated in the block? (d) If the bullet stops in a distance of 4.00 cm inside the block, estimate the force of friction. SOLUTION : (b) The problem consists of two stages. The ﬁrst stage is from before to after the impact. The second stage is from just after the impact to the ﬁnal position. In the ﬁrst stage, the linear momentum is conserved. In the second stage, the energy is conserved. Let us denote the initial speed of the bullet as v0 . So: pbefore = pafter mv0 = (m + M ) v (4) M +m v0 m M y=0 Then using the conservation of energy of the combined mass rising to height h after the impact: 1 (M + m) v 2 = (M + m) gh 2 ⇒ v = 2gh (a) L L y=h (5) Substituting v from eq. 5 into eq. 4 and solving for v0 : v0 = m+M m 2gh = 2.00 kg + 0.0100 kg 0.0100 kg 2 9.80 m/s2 (0.0500 m) = 199 m/s (c) Assuming no sound generated in the impact, all the energy lost will turn into heat energy. The heat energy generated is equal to the diﬀerence between the energy before the impact

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