1) A very strong boy throws a wad of clay with mass m. = 0.2 kg at a speed of v. = 30 m/s at a stationary toy train car sitting on a horizontal section of a toy train track, as shown in the diagram (which is not drawn to scale). The clay sticks to the back of the train car and remains stuck to the train following the collision. The train car has mass m,= 0.4 kg. You may neglect friction of the wheels and track. Throwing clay at a toy train car. a) What is the speed of the train car immediately after it is struck by the clay? b) What fraction of the clay's initial kinetic energy was lost to forms of energy other than kinetic energy as a result of the collision? c) The moving train car then ascends a 30° incline that rises a total of Ah = 15/4 m (3.75 m) above the lower section of the track. What is the speed of the train car once it reaches the top of this incline? d) How long does it take the train car to ascend the 30° incline? e) The train car then runs into a spring with total length L = 0.8 m and spring constant k = 60 N/m. What is the length of the compressed spring at the moment the train car comes to a stop?
1) A very strong boy throws a wad of clay with mass m. = 0.2 kg at a speed of v. = 30 m/s at a stationary toy train car sitting on a horizontal section of a toy train track, as shown in the diagram (which is not drawn to scale). The clay sticks to the back of the train car and remains stuck to the train following the collision. The train car has mass m,= 0.4 kg. You may neglect friction of the wheels and track. Throwing clay at a toy train car. a) What is the speed of the train car immediately after it is struck by the clay? b) What fraction of the clay's initial kinetic energy was lost to forms of energy other than kinetic energy as a result of the collision? c) The moving train car then ascends a 30° incline that rises a total of Ah = 15/4 m (3.75 m) above the lower section of the track. What is the speed of the train car once it reaches the top of this incline? d) How long does it take the train car to ascend the 30° incline? e) The train car then runs into a spring with total length L = 0.8 m and spring constant k = 60 N/m. What is the length of the compressed spring at the moment the train car comes to a stop?
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please answer remaining parts with reasoning and steps even if minimal step
Transcribed Image Text:1)
A very strong boy throws a wad of clay with mass m. = 0.2 kg at a speed of v. = 30 m/s at a
stationary toy train car sitting on a horizontal section of a toy train track, as shown in the diagram
(which is not drawn to scale). The clay sticks to the back of the train car and remains stuck to
the train following the collision. The train car has mass m,= 0.4 kg. You may neglect friction of
the wheels and track.
Throwing clay at a toy train car.
a) What is the speed of the train car immediately after it is struck by the clay?
b) What fraction of the clay's initial kinetic energy was lost to forms of energy other than kinetic
energy as a result of the collision?
c) The moving train car then ascends a 30° incline that rises a total of Ah = 15/4 m (3.75 m)
above the lower section of the track. What is the speed of the train car once it reaches the top
of this incline?
d) How long does it take the train car to ascend the 30° incline?
e) The train car then runs into a spring with total length L = 0.8 m and spring constant
k= 60 N/m. What is the length of the compressed spring at the moment the train car comes to
a stop?
spring e
train car
clay
frack
300
Transcribed Image Text:I nsing couservation of momentum.
me Ve= (mpt me) Vt
Given
me= 0.2 Kg
me =0.4 kg
= 3om Is
0.2 X 30= O.6x Vt
>
V= lom/s
O Now enegy of lay mev= x 0.2X g00
f mcv= Jx 0-2X go0
う Kc- 90J
energy of clay + train) = f xo.¢ x100 $ 30
3 of clay's energy was used to de form
the clay, hat energy, Bomnd energy
(e) accelention is along inclined
plane dowmwards
44
Now hsing ve u+ 2as
masings
ng
a = - gsino
&こ A4
Sino
v= 100 - 2X 10X 3.75
V= 5 m/s
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