(1)
A boy throws a ball on a wall with different velodties and the ball bounces back in 0.1
second as shown in the figure. Observe the plctures carefully and answer questions below
Case 1.
ball is moving with
10m/s velocity to
wall and comes back
to 5m/s
Case 2.
In case 2 ball is moving with 30m/s velocity to the wall and comes back to 20m/s.
10.1) If the same ball is used in both the cases, In which case the force exerted by the wall
on the ball would be greater? Justify your answer with relevant explanation.
Answers
Answer:
The laws of reflections are:
1) The incident ray, the normal and the reflected ray, all lie in the same plane.
2) The angle of incidence is equal to the angle of reflection.
Explanation:
Answer:
Explanation:
Let us assume the boy throws the ball at a Horizontal speed u and a vertical speed v.
Let the distance where the ball bounces off the ground be d.
g=10m/s
2
Time is taken by the ball to hit the wall =
u
6
In this time, Δy=3−1.4=1.6m
Using Kinematics relations, 1.6=6
u
v
−5(
u
6
)
2
---------(1)
As the collision is elastic at the wall, horizontal speed changes its sign, vertical speed remains the same.
Time taken by the ball to reach a height of 1.4m is
u
12−2d
From time of flight formula,
5
v
=
u
12−2d
-------------------(2)
At the height of 1.4m the vertical speed of the ball is −v and time taken to reach the ground is
u
d
Applying kinematics relations in the y direction,
−1.4=−v u d
−5( ud ) 2 -------------------(3)
Substituting (2) in (1) and (3), we have
1.6=
u
2
6(60−10d)−180
⇒1.6u
2
=180−60d ----------(4)
and 8.4u
2
=(1.6u
2
+180)d+30d
2 ------------(5)
Substituting (4) in (5)
30d
2 −675d+945=0⇒d=1.5m or d=21m
As d=21m is not possible, d=1.5m