Two towers AB and CD are d apart as shown. A ball of mass m is thrown from AB horizontally with a velocity of 10 m/s Simultaneously another ball of mass 2m is thrown from CD at 37 with the vertical (as shown) with initial velocity of 10 m/s as well. The two bals move in the same vertical plane, collide in mid air Calculate the distance d between the towers
Answers
Explanation:
Method 1:
a. Let t be the time taken for collision. For mass m thrown horizontally from A for horizontal motion,
PM=10t ...(i)
For vortical motion, u
y
=0,s
y
=y,a
y
=g
S
y
=u
y
t+
2
1
a
y
t
2
⇒Y=
2
1
gt
2
....(ii)
v
y
=u
y
+a
y
t=gt ....(iii)
For mass 2m thrown from C, for horizontal motion
QM=[10cos60
o
]t⇒QM=5t .....(iv)
For vertical motion: u
y
=10sin60
o
=5
3
⇒v
y
=5
3
+gt .....(v)
a
y
=g,S
y
=y+10,S=ut+
2
1
at
2
⇒y+10=5
3
t+
2
1
gt
2
.....(vi)
From (i) and (vi),
2
1
gt
2
+10=5
3
t+
2
1
gt
2
⇒t=
3
2
s
BD=PM+MQ
=10t+5t
=15t=15×
3
2
=10
3
=17.32m
b. Applying the conservation of linear momentum (during collision of the masses at M) in the horizontal direction, we have
m×10−2m10cos60
o
=3m×v
x
⇒10m−10m=3m×v
x
⇒v
x
=0
Since the horizontal momentum comes out to be zero, the combination of masses will drop vertically downwards and fall at E,BE=PM=10t=10×
3
3
=11.547m
Method 2:
Acceleration of A and Cboth is 9.8ms
−1
downward. Therefore, the relative acceleration between them is zero, i.e., the relative motion between them will be a straight line.
Now assuming A to be at rest, the condition of collision will be that
v
CA
=
v
C
−
v
A
= relative velocity of C w.r.t. A should be along CA.
v
A
=10
i
^
v
B
=−5
i
^
−5
3
j
^
v
BA
=−5
i
^
−5
3
j
^
−10
i
^
∴
v
BA
=−15
i
^
−5
3
j
^
∴tan60
o
=
10
d
ord=10
3
m
solution
Answered By
toppr
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