Q. 01. A ball hits a wall horizontally at 6m/s. It rebounds horizontally at 4.4m/s. The ball is in contact with the wall for 0.04s. What is the acceleration of the ball?
Initial velocity(u) = 6.0m/s
Final velocity(v) = -4.4 m/s (because direction of ball has become opposite)
time = 0.04 s
thus, acceleration (a) = (v-u)/t
a= [-4.4 – 6.0]/ 0.04
after multiplying -10.4/0.04 by 100/100 (to make calculations simpler), we get
a= -1040/4 = -260 m/s2
Q.02. A cheetah is the fastest land animal and it can achieve a peak velocity of 100 km per hour up to distances less than 500 metres. If the cheetah spots his prey at a distance of 100 meters what is the minimum time it will take to get its prey?
If the cheetah spots the prey at its top speed, the cheetah will hunt down the prey with the speed of 100 km/h. = 27.7 m/s
Now, time = d/s = 100/27.7 = 3.6 sec.
So, the minimum time the cheetah will take to get the prey is 3.6 s.
Q.03. A police jeep is chasing velocity of 45 km/h. A thief in another jeep moving with a velocity of 153 km/h. police fires a bullet with muzzle velocity of 180 m/s the velocity it will strike the car of the thief is?
Given: Velocity of police jeep = 45 m/hr = 12.5 m/s Velocity of thief’s jeep = 153 km/hr = 42.5 m/s
Velocity of bullet = 180 m/s
Now, since bullet is fired from police jeep which is going at 42.5 m/s,
Therefore, velocity of bullet is (180 + 12.5) = 192. 5 m/s.
To calculate velocity with which bullet with hit the thief we use concept of relative velocity.
Therefore, we have VBT = VB - VT
Here VB is velocity of bullet and VT is velocity of thief
VBT = 192. 5m/s - 42.5 m/s = 150m/s
Q. 05. A car is travelling at 20 m/s along a road. A child runs out into the road 50 m ahead and the car driver steps on the brake pedal. What must the car’s deceleration be if the car is to stop just before it reaches the child?
u = 20m/s ; s = 50m and v = 0m/s
Using, V^2 = u^2 + 2as
a = (v^2 – u^2)/2s
a = (0^2 - 20^2)/(2 x 50) = - 4m/s^2
Acceleration = -4 m/s2 or, retardation = 4 m/s2
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