M2 qs again

This is an M2 qs from the Edexcel textbk—pg 28, qs8

A particle of mass .5 kg is at rest on a horizontal table. It receives a blow of impulse 2.5Ns.
a. calculate the speed whith which P is moving immediately after the blow.

The height of the table is .9m and the floor is horizontal. In an initial model of the situation the table is assumed to b smooth.
b. calculate the horizontal dist from the edge of the table to the point where P hits the ground .

in a refinement model of the situation the table is assumed to b rough. The coefficient of friction between P and the table is .2
c. calculate the deceleration of P.

given that P travels .4m to the edge of the table,
d. calculate the time which elapses btwwen P receiving blow to P hiting the floor.

Ive done the first part like this:

Impulse= change in momentum= m(v-u)
2.5=.5(v-0)
Therefore v=5 m/s(correct answer)

Can someone plz help me out with the other parts, I don’t know how to do them

thanks

Reply 1

dvs

10

b) Think of it as projectile motion where the intial speed is 5 m/s, and there is no angle of elevation.

c) Newton's second law will do: friction = uR = 0.2(mg) = - ma. So the deceleration is -a = 0.2g = 1.96 m/s^2.

d) Use the regular kinematic equations to find the time until P reaches the edge, then proceed like before by considering projectile motion.

Reply 2

sandra

OP

1

dvs

b) Think of it as projectile motion where the intial speed is 5 m/s, and there is no angle of elevation.

c) Newton's second law will do: friction = uR = 0.2(mg) = - ma. So the deceleration is -a = 0.2g = 1.96 m/s^2.

d) Use the regular kinematic equations to find the time until P reaches the edge, then proceed like before by considering projectile motion.

i tried doing the b part now using
S=ut +half at square
whre S= -.9,u=5 , a = +9.8

but im getting the wrong answer :frown:

cud u plz help?

Reply 3

Malik

12

b)

s=ut+0.5at² (theres no vertical motion so u=0)
0.9=4.9t²
t=3/7

Horizontal distance=st= 5(3/7)=15/7

c)
F=mueR
mue=0.2 and R=0.5g
F=0.98

F=ma
0.98=0.5a
a=1.96

d)
s=d/t
5=0.4/t
t=0.08s

We calculated before time hit floor=3/7
so total time=0.08+3/7= 89/175s

Reply 4

sandra

OP

1

Malik

b)

s=ut+0.5at² (theres no vertical motion so u=0)
0.9=4.9t²
t=3/7

Horizontal distance=st= 5(3/7)=15/7

c)
F=mueR
mue=0.2 and R=0.5g
F=0.98

F=ma
0.98=0.5a
a=1.96

d)
s=d/t
5=0.4/t
t=0.08s

We calculated before time hit floor=3/7
so total time=0.08+3/7= 89/175s