# Physics 9702 Doubts | Help Page 31

__Question 193: [Dynamics > Momentum]__Two train carriages each of mass 5000 kg roll toward one another on level track. One is travelling at 2.00 m s

^{–1}and other at 1.00 m s

^{–1}, as shown.

They collide and join together.

What is the kinetic energy lost during collision?

A 1250 J B 7500 J C 11 250 J D 12 500 J

**Reference:**

*Past Exam Paper – June 2014 Paper 12 Q7*

__Solution 193:__**Answer: C.**

Total Kinetic Energy before
collision = ½ (5000)(2)

^{2}+ ½ (5000)(1)^{2}= 12500J
From the conservation of
momentum, the sum of momentum before collision is equal to the sum of momentum
after collision.

Let the final speed after collision
= v

Take the direction towards the right
as the positive direction.

5000(2) + 5000(-1) = 10000v

Speed, v = 5000/10000 = 0.5ms

^{-1}
Kinetic Energy after collision = ½
(10000)(0.5)

^{2}= 1250J
Kinetic Energy lost = 12500 – 1250 =
11 250J

__Question 194: [Kinematics > Linear motion > Graph]__
In an experiment to determine
acceleration of free fall using a falling body, what would lead to value that
is too large?

A air resistance

B dimensions of body are too large

C measured distance longer than true
distance

D measured time longer than true
time

**Reference:**

*Past Exam Paper – June 2012 Paper 11 Q6*

__Solution 194:__**Answer: C.**

The measurements that can
be taken for a falling body are the distance between 2 points, and the time
that the falling body takes to travel between the 2 points. From these
measurements, a graph of distance against (time)

^{2}may be plotted to obtain the acceleration of free fall, for example.
Air resistance acts against motion,
and thus causes a resultant acceleration on the body which is less than the
acceleration of free fall.

Dimension of the body would tell
about the size (e.g. volume, …) of the body. This information is not necessary
for the determination of the acceleration of free fall.

As explained above, the acceleration
may be obtained by the gradient of a distance – (time)

^{2}graph. Gradient = distance / (time)^{2}. So, having a longer measured distance than the true distance causes the gradient to is larger, while having a longer measured time than the true time causes the gradient to be smaller.

__Question 195: [Kinematics & Dynamics]__
Bullet of mass 2.0 g is fired
horizontally into block of wood of mass 600 g. Block is suspended from strings
so that it is free to move in vertical plane. Bullet buries itself in block. Block
and bullet rise together through vertical distance of 8.6 cm, as shown.

**(a)**

(i) Calculate change in
gravitational potential energy, Î”E

_{p}of block and bullet
(ii) Show that initial speed of
block and bullet, after they began to move off together, was 1.3ms

^{–1}**(b)**Using information in (a)(ii) and principle of conservation of momentum, determine speed of bullet before impact with block

**(c)**

(i) Calculate kinetic energy of
bullet just before impact

(ii) Explain what can be deduced
from answers to (c)(i) and (a)(i) about type of collision between bullet and
block

**Reference:**

*Past Exam Paper – June 2005 Paper 2 Q3*

__Solution 195:__**(a)**

(i)

{Total mass = mass of
bullet + mass of block = 2 + 600 = 602g = 0.602g}

Î”E

_{p}= mgÎ”h = 0.602 x 9.8 x 0.086 = 0.51J
(ii)

{v

^{2}= u^{2}+ 2as = 0 + 2gh. For the OR case, KE of bullet and block = Change in gravitational PE. ½ mv^{2}= 0.51J}
v

^{2}(= 2gh) = 2 (9.8) (0.086)__or__v^{2}= (2 x 0.51) / 0.602
Initial speed of block and bullet, v
= 1.3ms

^{-1}**(b)**

{Sum of momentum before
collision = Sum of momentum after collision}

2v = 602 x 1.3

Speed of bullet, v = 390ms

^{-1}**(c)**

(i)

Kinetic energy of bullet, E

_{k}= ½mv^{2}= ½ (0.002) (390^{2}) = 152J or 153J or 150J
(ii)

The kinetic energy is not the same /
changes OR the kinetic energy before the impact is greater than the kinetic
energy / potential energy after the impact. So, the collision must be
inelastic.

__Question 196: [Forces]__Car of mass m travels at constant speed up a slope at an angle Î¸ to horizontal, as shown in the diagram. Air resistance and friction provide resistive force F.

What force is needed to propel car at this constant speed?

A mg cos Î¸

B mg sin Î¸

C mg cos Î¸ + F

D mg sin Î¸ + F

**Reference:**

*Past Exam Paper – June 2012 Paper 11 Q14*

__Solution 196:__**Answer: D.**

**direction as the resistive force F). This component is equal to mg sinÎ¸.**

__same__For the car to move at

__constant speed__, the force required to move the car (up) along the slope should be equal to the sum of the forces acting (down) along the slope such that the resultant force and hence

__resultant acceleration is zero__.

Thus, the force = mg sinÎ¸ + F

in solution 193 how is 5000(2)+5000(-1)=10000???

ReplyDeleteits 10000v

Deletewe are equating momentum. it's an equation

can someone explain question 195"b"

ReplyDeleteit seems to be self-explicit. What is your confusion?

ReplyDeleteIn solution 193 how v=5000/10000

ReplyDeletejust evaluate the left hand side of the equation above it, then make v subject of formula

Delete